Week 4: Differential Equations & Linear Algebra Notes
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Christian Parkinson GRE Prep: Diff. Eq. & Lin. Alg. Notes 1 Week 4: Differential Equations & Linear Algebra Notes After calculus, the most common topic on the math subject GRE is linear algebra, foll- wed by differential equations. Since the latter is a more natural continuation of calculus (and since it will take much less time), I cover it first. Differential Equations Definition 1 (Differential Equation). An ordinary differential equation is an equation of the form F (x; y; y0; y00; : : : ; y(n)) = 0 for some function F : Rn+2 ! R where y(k) denotes the kth derivative of y with respect to the independent variable x. The order of the differential equation is equal to the highest order derivative of u which appears; the above equation is nth order (assuming F does actually depend on its last variable). Note, we often replace y with u which can be thought to stand for \unknown." The goal then, is to find the unknown function which satisfies the equation. On the math subject GRE, questions are mostly limited to equations where the highest order derivative of the unknown function can be isolated; that is, equations of the form y(n) = f(x; y0; : : : ; y(n−1)): Many differential equations can be solved by simply recalling facts from calculus. Example 2. Solve the differential equations (a) u0 = u; (b) y00 = −4y: Solution. For (a), we can phrase the differential equation as a question: \What function will remain the same upon differentiating?" We recall from calculus that u(x) = ex is the function which is invariant under differentiation. Indeed, this a solution to the equation in (a). For (b), asking a similar question, we reason that y should be a function which, upon twice differentiating, returns the negative of itself (with a constant multiple). You may recall from calculus that the function sin(x) returns its negative after twice differentiating and then to account for the constant multiple, you can see that y(x) = sin(2x) is a solution. A natural question would be to ask if we have found all solutions to a given equation. Of course, in the above example, we did not. For (a), any constant times ex would work just as well so a full family of solutions would be u(x) = Cex for some arbitrary constant C 2 R. For (b), besides simply mutiplying by a constant, we see y(x) = cos(2x) works as well. Indeed, combinations of the two also work so the full family of solutions is give by y(x) = A sin(2x) + B cos(2x) for some arbitrary constants A; B 2 R. Both these equations have special structure which is defined here. Christian Parkinson GRE Prep: Diff. Eq. & Lin. Alg. Notes 2 Definition 3 (Linearity). A differential equation F (x; y; y0; : : : ; y(n)) = 0 said to be linear if the function F is linear in the arguments (y; y0; : : : ; y(n)); that is, F (x; αy + z; αy0 + z0; : : : ; αy(n) + z(n)) = αF (x; y; y0; : : : ; y(n)) + F (x; z; z0; : : : ; z(n)) for all n times differentiable y; z : R ! R and all α 2 R. In the terminology of linear algebra, a differential equation is linear if its solution set forms a (possibly affine) subspace of the vector space of continuous functions on R. Intuitively, a differential equation is linear if the unknown function and its derivatives do not appear in any non-linear way. In symbols, a linear nth order differential equation has the form (n) 00 0 an(x)y + ··· + a2(x)y + a1(x)y + a0(x)y = f(x) for some functions ak; f : R ! R (k = 0; : : : ; n). Typically, we also require that an(x) 6= 0 for any x, so that we may divide by an(x) and (n) eliminate the coefficient of y . If an(x) = 0 at some x, this x is called a singular point of the differential equation. Solving differential equations with singular points can be quite difficult and is beyond the scope of these notes. Linear equations are always solvable analytically. We build toward the general solution of a first-order linear equation in a few steps. Definition 4 (Separability). A first-order differential equation is called separable if it is of the form f(x) y0 = : g(y) That is, separable equations have the variables on the right hand side separated into two different functions. Proposition 5 (Solution to Separable Equations). The solution to f(x) y0 = g(y) is given by the pair of integrals Z Z g(y)dy = f(x)dx: Formally, we can arrive at this by manipulating differentials: dy f(x) Z Z = =) g(y)dy = f(x)dx =) g(y)dy = f(x)dx: dx g(y) Christian Parkinson GRE Prep: Diff. Eq. & Lin. Alg. Notes 3 Of course, this actually depends on the chain rule. Example 6. Solve the differential equation y0 = (1 + y2)(4x3 + 2x): Solution. We see that Z dy Z = (4x3 + 2x2)dx =) arctan y = x4 + x2 + C =) y = tan(x4 + x2 + C): 1 + y2 It can be readily checked that this function does satisfy the differential equation. From this, we can easily solve any equation of the form y0 = p(x)y: We see that such an equation has the solution R y(x) = e p(x)dx: This observation gives us a method for solving any first order linear equation. Proposition 7 (Method of Integrating Factor). The solution to the first-order linear equation y0 + p(x)y = q(x) is given by R Z R y(x) = e− p(x)dx e p(x)dxq(x)dx : This formula deserves some motivation. In line with separable equations (or even simpler equations of the form y0(x) = f(x)), we see that solving a differential equation is akin to \integrating the equation." With this in mind, it would be ideal if the left hand side of the above equation was a perfect derivative so we could just integrate it away. In general, there will not be Y (x) such that Y 0(x) = y0(x) + p(x)y; i.e., the left hand side will not be a perfect derivative. However, we can fix that by cleverly multiplying by some other function µ(x) which we call an integrating factor. Indeed, our new equation will read µ(x)y0(x) + µ(x)p(x)y(x) = µ(x)q(x): Now, if µ0 = p(x)µ, then the left hand side will read µ(x)y0(x) + µ0(x)y(x) which is the derivative of µ(x)y(x). Thus we simply choose µ satisfying µ0 = p(x)µ. However, R from the above, we know that µ(x) = e p(x)dx satisfies that equation. Thus we see, R R R d h R i R e p(x)dxy0 + p(x)e p(x)dxy = e p(x)dxq(x) =) e p(x)dxy(x) = e p(x)dxq(x) dx Christian Parkinson GRE Prep: Diff. Eq. & Lin. Alg. Notes 4 whence integrating gives R Z R R Z R e p(x)dxy(x) = e p(x)dxq(x)dx =) y(x) = e− p(x)dx e p(x)dxq(x)dx which is the solution listed above. Example 8. Solve the equation 1 y0(x) + y(x) = cos(x2): x R Solution. According the formula, we should multiply the equation by µ(x) = e (1=x)dx = eln(x) = x: Doing this gives d xy0(x) + y(x) = x cos(x2) =) (xy(x)) = x cos(x2): dx Integrating yields sin(x2) sin(x2) C xy(x) = + C =) y(x) = + 2 2x x where C is an arbitrary constant. It may irk you that all of these solutions have this lingering arbitrary constant C. This has something to do with the fact that differentiation eliminates constants so when we solve the equation (think of \integrating the equation") we need to reintroduce constants that may have been eliminated. Often times, along with a differential equation, a given value y(x0) = y0 is specified. This value can be used to solve for the constant C. For example,p along with the above equation, the question may have specifiedp that y should satisfy y( π) = 1: We found our general solution, so substituting x = π gives p sin(π) C y( π) = p + p : 2 π π We are told this value should be 1, so this gives C p p = 1 =) C = π: π p Thus the function y which solves the equation and satisfies y( π) = 1 is given by p sin(x2) π y(x) = + : 2x x Such questions which give an equation and specify a value for the unknown function at a point are called Initial Value Problems or Boundary Value Problems depending on the con- text (and the way in which the values of the unknown function are specified). We won't Christian Parkinson GRE Prep: Diff. Eq. & Lin. Alg. Notes 5 address this point any further. This gives a solution to all linear first order equations. With separation of variables we can also solve some non-linear equations. Most non-linear equations will not be solvable an- alytically. However, without solving we can still determine some of the behavior for solutions to certain equations. Definition 9 (Autonomous Equations). A first-order differential equation is called autonomous if it is of the form y0 = f(y): Note that all autonomous equations are separable and thus can, in principle, be solved by integration. However, that integration could be too difficult to be feasibly performed.