Overview An Example Double Check Further Discussion
Variation of Parameters
Bernd Schroder¨
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters y = yp + yh 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation. 3. The particular solution can be obtained as follows. 3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation. 3. The particular solution can be obtained as follows. 3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp + yh
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3. The particular solution can be obtained as follows. 3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp + yh 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp + yh 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation. 3. The particular solution can be obtained as follows.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp + yh 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation. 3. The particular solution can be obtained as follows. 3.1 Assume that the parameters in the solution of the homogeneous equation are functions.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp + yh 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation. 3. The particular solution can be obtained as follows. 3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Variation of Parameters 1. The general solution of an inhomogeneous linear differential equation is the sum of a particular solution of the inhomogeneous equation and the general solution of the corresponding homogeneous equation. y = yp + yh 2. Variation of Parameters is a way to obtain a particular solution of the inhomogeneous equation. 3. The particular solution can be obtained as follows. 3.1 Assume that the parameters in the solution of the homogeneous equation are functions. (Hence the name.) 3.2 Substitute the expression into the inhomogeneous equation and solve for the parameters.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z x 1 f (t) Z x 1 f (t) yp(x) = −y1(x) y2(t) dt + y2(x) y1(t) dt, x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
where y1, y2 are linearly independent solutions of the 0 0 homogeneous equation and W(y1,y2) := y1y2 − y2y1. W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as y1 y2 W(y1,y2) = det 0 0 . y1 y2 That’s it.
Overview An Example Double Check Further Discussion
Variation of Parameters For second order equations, we obtain the general formula
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters where y1, y2 are linearly independent solutions of the 0 0 homogeneous equation and W(y1,y2) := y1y2 − y2y1. W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as y1 y2 W(y1,y2) = det 0 0 . y1 y2 That’s it.
Overview An Example Double Check Further Discussion
Variation of Parameters For second order equations, we obtain the general formula
Z x 1 f (t) Z x 1 f (t) yp(x) = −y1(x) y2(t) dt + y2(x) y1(t) dt, x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as y1 y2 W(y1,y2) = det 0 0 . y1 y2 That’s it.
Overview An Example Double Check Further Discussion
Variation of Parameters For second order equations, we obtain the general formula
Z x 1 f (t) Z x 1 f (t) yp(x) = −y1(x) y2(t) dt + y2(x) y1(t) dt, x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
where y1, y2 are linearly independent solutions of the 0 0 homogeneous equation and W(y1,y2) := y1y2 − y2y1.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters The Wronskian can also be represented as y1 y2 W(y1,y2) = det 0 0 . y1 y2 That’s it.
Overview An Example Double Check Further Discussion
Variation of Parameters For second order equations, we obtain the general formula
Z x 1 f (t) Z x 1 f (t) yp(x) = −y1(x) y2(t) dt + y2(x) y1(t) dt, x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
where y1, y2 are linearly independent solutions of the 0 0 homogeneous equation and W(y1,y2) := y1y2 − y2y1. W is also called the Wronskian of y1 and y2.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters That’s it.
Overview An Example Double Check Further Discussion
Variation of Parameters For second order equations, we obtain the general formula
Z x 1 f (t) Z x 1 f (t) yp(x) = −y1(x) y2(t) dt + y2(x) y1(t) dt, x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
where y1, y2 are linearly independent solutions of the 0 0 homogeneous equation and W(y1,y2) := y1y2 − y2y1. W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as y1 y2 W(y1,y2) = det 0 0 . y1 y2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Variation of Parameters For second order equations, we obtain the general formula
Z x 1 f (t) Z x 1 f (t) yp(x) = −y1(x) y2(t) dt + y2(x) y1(t) dt, x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
where y1, y2 are linearly independent solutions of the 0 0 homogeneous equation and W(y1,y2) := y1y2 − y2y1. W is also called the Wronskian of y1 and y2. The Wronskian can also be represented as y1 y2 W(y1,y2) = det 0 0 . y1 y2 That’s it.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Solution of the homogeneous equation. y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = = −2 1,2 2
−2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = = −2 1,2 2
−2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters λ 2eλx + 4λeλx + 4eλx = 0 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = = −2 1,2 2
−2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation. y00 + 4y0 + 4y = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = = −2 1,2 2
−2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation. y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters √ −4 ± 16 − 16 λ = = −2 1,2 2
−2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation. y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0 λ 2 + 4λ + 4 = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters = −2
−2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation. y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = 1,2 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters −2x −2x yh = c1e + c2xe
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation. y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = = −2 1,2 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Solution of the homogeneous equation. y00 + 4y0 + 4y = 0 λ 2eλx + 4λeλx + 4eλx = 0 2 λ + 4λ + 4 = 0 √ −4 ± 16 − 16 λ = = −2 1,2 2
−2x −2x yh = c1e + c2xe
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Computing the Wronskian. −2t −2t −2t −2t −2t W(y1,y2)(t) = e e − 2te − te −2e = e−4t − 2te−4t + 2te−4t = e−4t
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters = e−4t − 2te−4t + 2te−4t = e−4t
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the Wronskian. −2t −2t −2t −2t −2t W(y1,y2)(t) = e e − 2te − te −2e
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters = e−4t
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the Wronskian. −2t −2t −2t −2t −2t W(y1,y2)(t) = e e − 2te − te −2e = e−4t − 2te−4t + 2te−4t
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the Wronskian. −2t −2t −2t −2t −2t W(y1,y2)(t) = e e − 2te − te −2e = e−4t − 2te−4t + 2te−4t = e−4t
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Computing the integrals. Z Z 1 f (t) 1 sin(t) −2t y1(t) dt = −4t e dt W(y1,y2)(t) a2(t) e 1 Z = e2t sin(t) dt
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z 1 sin(t) = e−2t dt e−4t 1 Z = e2t sin(t) dt
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y1(t) dt W(y1,y2)(t) a2(t)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z = e2t sin(t) dt
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z Z 1 f (t) 1 sin(t) −2t y1(t) dt = −4t e dt W(y1,y2)(t) a2(t) e 1
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z Z 1 f (t) 1 sin(t) −2t y1(t) dt = −4t e dt W(y1,y2)(t) a2(t) e 1 Z = e2t sin(t) dt
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Computing the integrals. Z 1 Z 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) dt 2 2 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4 Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 Z 1 = e2t sin(t) − e2t cos(t) dt 2 2 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4 Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z e2t sin(t) dt
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4 Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) dt 2 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4 Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) dt 2 2 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4 Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) dt 2 2 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) dt 2 2 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) dt 2 2 1 1 1 Z 1 = e2t sin(t) − e2t cos(t) − e2t − sin(t) dt 2 2 2 2 1 1 1 Z = e2t sin(t) − e2t cos(t) − e2t sin(t) dt 2 4 4 5 Z 1 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 4 2 4 Z 2 1 e2t sin(t) dt = e2t sin(t) − e2t cos(t) 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) Z 1 sin(t) = te−2t dt e−4t 1 Z = te2t sin(t) dt 2 1 Z 2 1 = t e2t sin(t) − e2t cos(t) − e2t sin(t) − e2t cos(t) dt 5 5 5 5 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z 1 sin(t) = te−2t dt e−4t 1 Z = te2t sin(t) dt 2 1 Z 2 1 = t e2t sin(t) − e2t cos(t) − e2t sin(t) − e2t cos(t) dt 5 5 5 5 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z = te2t sin(t) dt 2 1 Z 2 1 = t e2t sin(t) − e2t cos(t) − e2t sin(t) − e2t cos(t) dt 5 5 5 5 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) Z 1 sin(t) = te−2t dt e−4t 1
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 Z 2 1 = t e2t sin(t) − e2t cos(t) − e2t sin(t) − e2t cos(t) dt 5 5 5 5 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) Z 1 sin(t) = te−2t dt e−4t 1 Z = te2t sin(t) dt
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) Z 1 sin(t) = te−2t dt e−4t 1 Z = te2t sin(t) dt 2 1 Z 2 1 = t e2t sin(t) − e2t cos(t) − e2t sin(t) − e2t cos(t) dt 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) Z 1 sin(t) = te−2t dt e−4t 1 Z = te2t sin(t) dt 2 1 Z 2 1 = t e2t sin(t) − e2t cos(t) − e2t sin(t) − e2t cos(t) dt 5 5 5 5 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Computing the integrals. Z 1 Z 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) dt 2 2 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4 Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 Z 1 = e2t cos(t) + e2t sin(t) dt 2 2 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4 Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z e2t cos(t) dt
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4 Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) dt 2 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4 Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) dt 2 2 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4 Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) dt 2 2 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) dt 2 2 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 Z 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) dt 2 2 1 1 1 Z 1 = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 2 2 2 1 1 1 Z = e2t cos(t) + e2t sin(t) − e2t cos(t) dt 2 4 4 5 Z 1 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 4 2 4 Z 2 1 e2t cos(t) dt = e2t cos(t) + e2t sin(t) 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5 2 1 2 2 1 = te2t sin(t) − te2t cos(t) − e2t sin(t) − e2t cos(t) 5 5 5 5 5 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5 2 1 2 2 1 = te2t sin(t) − te2t cos(t) − e2t sin(t) − e2t cos(t) 5 5 5 5 5 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 2 2 1 = te2t sin(t) − te2t cos(t) − e2t sin(t) − e2t cos(t) 5 5 5 5 5 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 2 1 − e2t sin(t) − e2t cos(t) 5 5 5 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5 2 1 = te2t sin(t) − te2t cos(t) 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5 2 1 2 2 1 = te2t sin(t) − te2t cos(t) − e2t sin(t) − e2t cos(t) 5 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5 2 1 2 2 1 = te2t sin(t) − te2t cos(t) − e2t sin(t) − e2t cos(t) 5 5 5 5 5 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Computing the integrals. Z 1 f (t) y2(t) dt W(y1,y2)(t) a2(t) 2 1 2 Z 1 Z = te2t sin(t) − te2t cos(t) − e2t sin(t) dt + e2t cos(t) dt 5 5 5 5 2 1 2 2 1 = te2t sin(t) − te2t cos(t) − e2t sin(t) − e2t cos(t) 5 5 5 5 5 1 2 1 + e2t cos(t) + e2t sin(t) 5 5 5 2 1 3 4 = te2t sin(t) − te2t cos(t) − e2t sin(t) + e2t cos(t) 5 5 25 25 logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Stating the general solution. Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t) 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5 3 4 = sin(x) − cos(x) 25 25 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t) 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5 3 4 = sin(x) − cos(x) 25 25 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Stating the general solution.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5 3 4 = sin(x) − cos(x) 25 25 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Stating the general solution. Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5 3 4 = sin(x) − cos(x) 25 25 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Stating the general solution. Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t) 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3 4 = sin(x) − cos(x) 25 25 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Stating the general solution. Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t) 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Stating the general solution. Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t) 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5 3 4 = sin(x) − cos(x) 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Stating the general solution. Z x 1 f (t) Z x 1 f (t) yp = −y1(x) y2(t) dt + y2(x) y1(t) dt x0 W(y1,y2)(t) a2(t) x0 W(y1,y2)(t) a2(t) 2 1 3 4 = −e−2x xe2x sin(x) − xe2x cos(x) − e2x sin(x) + e2x cos(x) 5 5 25 25 2 1 +xe−2x e2x sin(x) − e2x cos(x) 5 5 3 4 = sin(x) − cos(x) 25 25 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 29 = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 1 = y(0)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 29 , c = 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 1 = y(0) = − + c 25 1
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3 3 55 11 = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 0 = y0(0)
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3 55 11 , c = 2c − = = 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 0 = y0(0) = − 2c + c 25 1 2
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 55 11 = = 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 0 = y0(0) = − 2c + c , c = 2c − 25 1 2 2 1 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 11 = 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 0 = y0(0) = − 2c + c , c = 2c − = 25 1 2 2 1 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0 Finding c1, c2. 4 3 y = − cos(x) + sin(x) + c e−2x + c xe−2x 25 25 1 2 4 3 y0 = sin(x) + cos(x) − 2c e−2x + c e−2x − 2xe−2x 25 25 1 2 4 29 1 = y(0) = − + c , c = 25 1 1 25 3 3 55 11 0 = y0(0) = − 2c + c , c = 2c − = = 25 1 2 2 1 25 25 5 4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Solve the Initial Value Problem y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0
4 3 29 11 y = − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 12 16 3 + + − sin(x) 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 − + + cos(x) 25 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 44 88 44 + − + xe−2x =? sin(x) 5 5 5
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 + − + + − e−2x 25 25 5 25 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters =? sin(x)
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x 25 25 5 25 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 + − + + − e−2x+ − + xe−2x =? sin(x) 25 25 5 25 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 3 29 11 4 − cos(x) + sin(x) + e−2x + xe−2x 25 25 25 5 4 3 58 11 +4 sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 5 4 3 116 11 + cos(x) − sin(x) + e−2x + −4e−2x + 4xe−2x =? sin(x) 25 25 25 5 16 12 4 12 16 3 − + + cos(x) + + − sin(x) 25 25 25 25 25 25 116 232 44 116 44 44 88 44 √ + − + + − e−2x+ − + xe−2x = sin(x) 25 25 5 25 5 5 5 5
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters √ = 1 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 y(0) = − + 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters √
4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 y(0) = − + = 1 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters √ = 0 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 y0(0) = − + 25 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters √
Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 y0(0) = − + = 0 25 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Beyond a certain level of complexity, that really is the way to go.
Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
4 3 29 11 Does y = − cos(x) + sin(x) + e−2x + xe−2x Solve 25 25 25 5 y00 + 4y0 + 4y = sin(x), y(0) = 1, y0(0) = 0?
4 29 √ y(0) = − + = 1 25 25 4 3 58 55 y0(x) = sin(x) + cos(x) − e−2x + e−2x − 2xe−2x 25 25 25 25 3 58 55 √ y0(0) = − + = 0 25 25 25 Alternatively, use a computer to double check the result. Beyond a certain level of complexity, that really is the way to go.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Double Checking with a Computer Algebra System
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Double Checking with a Computer Algebra System
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Double Checking with a Computer Algebra System
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Double Checking with a Computer Algebra System
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Double Checking with a Computer Algebra System
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Double Checking with a Computer Algebra System
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients. 2. Integrals get challenging or even impossible. 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here. 4. Remember, the beauty is that we can get a solution. 5. The simple solutions of toy problems are best forgotten. 6. We now have a tool that can handle real life situations.
Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 2. Integrals get challenging or even impossible. 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here. 4. Remember, the beauty is that we can get a solution. 5. The simple solutions of toy problems are best forgotten. 6. We now have a tool that can handle real life situations.
Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here. 4. Remember, the beauty is that we can get a solution. 5. The simple solutions of toy problems are best forgotten. 6. We now have a tool that can handle real life situations.
Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients. 2. Integrals get challenging or even impossible.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 4. Remember, the beauty is that we can get a solution. 5. The simple solutions of toy problems are best forgotten. 6. We now have a tool that can handle real life situations.
Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients. 2. Integrals get challenging or even impossible. 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 5. The simple solutions of toy problems are best forgotten. 6. We now have a tool that can handle real life situations.
Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients. 2. Integrals get challenging or even impossible. 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here. 4. Remember, the beauty is that we can get a solution.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters 6. We now have a tool that can handle real life situations.
Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients. 2. Integrals get challenging or even impossible. 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here. 4. Remember, the beauty is that we can get a solution. 5. The simple solutions of toy problems are best forgotten.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters Overview An Example Double Check Further Discussion
Further Remarks on Variation of Parameters 1. The method can be applied to higher order equations, systems, and equations with non-constant coefficients. 2. Integrals get challenging or even impossible. 3. So the Derivative Form of the Fundamental Theorem of Calculus and numerical integration are frequently used here. 4. Remember, the beauty is that we can get a solution. 5. The simple solutions of toy problems are best forgotten. 6. We now have a tool that can handle real life situations.
logo1 Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science Variation of Parameters