Tutorial Problems #5 MAT 267 – Advanced Ordinary Di↵Erential Equations

Tutorial Problems #5 MAT 267 – Advanced Ordinary Di↵erential Equations – Winter 2016 Christopher J. Adkins

Solutions

d Reduction of Order Via Di↵erential Operators Let D = dx be our di↵erential operator. Then any n-th order linear non-homogeneous equation may be written as

n n n 1 n d L(D)[y(x)] = f(x)whereL(D)=D + an 1D + ...+ a0 D = dxn ✓ ◆ with ai R. Factor L into a product of it’s roots (which may be complex and we’ll deal with later), i.e. 2 L(D)=(D ) ...(D ) 1 n

Notice this factorization is not possible if ai are functions since the di↵erential operator isn’t commutative (D D = D D ). Thus, if we let y =(D )y and y =(D )y we e↵ectively reduce the n-th order 1 2 2 1 n n i i i+1 equations into n ﬁrst order equations (which we know how to handle)

pg. 267 - # 28 Solve (using reduction of order)

2 y00 + y0 = x +2x

Solution We see that if L = D2 + D,then

L(D)[y(x)] = x2 +2x is the ODE we’re looking to solve. Notice we may use the above method to deduce

2 L(D)=D(D + 1) = u0 = x +2x where (D + 1)y = u ) The above ODE in u is separable, thus

3 2 x 2 u(x)= x +2xdx = + x + C1 C1 R 3 2 Z We now know 3 x 2 y0 + y = + x + C 3 1 which is a ﬁrst order linear ODE, we know this may be solved using an integrating factor. We know 1 y(x)= µ(x)g(x)dx where µ(x)=exp dx = ex µ(x) Z ✓Z ◆

1 Tutorial #5 – Winter 2016 MAT 267

Thus the general solution to the ODE is

3 3 x x x 2 x x y(x)=e e + x + C1 dx = + C2e + C1 C2 R 3 3 2 Z ✓ ◆

1 The Inverse of a Di↵erential Operator Let’s talk about D now. Formally we need an operator with 1 the property if Dx = y,thenx = D y. Intuitively, you should think the integral operator is a natural left inverse for D since d f(x)dx = f(x) dx Z by the fundamental theorem of calculus. Now what about factors of (D ) we had...using a formal series expansion(notably a geometric series), we may algebraically write

2 3 1 1 1 D D D (D ) = = 1+ + + ... (1 D/) 2 3  Convergence of this series is a slight issue at the moment...but for any solution that terminates after a ﬁnite number of derivatives we know convergence is guaranteed. Let’s revisit the example we just saw.

pg. 267 - # 28 Solve (using Inverse Operators)

2 y00 + y0 = x +2x

Solution As we saw before we have 1 D(D + 1)y = x2 +2x = y (x)= (x2 +2x) ) p D(D + 1)

Notice we’ll only be able to pick up the particular solution to the ODE with this method (not the general) since

L is not injective in general(i.e. L[yhom] = 0). Expanding the inverse into formal series shows

1 1 y (x)= 1+D + D2 (x2 +2x)= 1+D (x2 +2x) p D D  Thus d x3 y (x)= (x2 +2x)dx (x2 +2x)+ (x2 +2x)= +2 p dx 3 Z You may recover the general solution using your knowledge of homogeneous equation, but seeing the eigenvalues x of = 0 and = 1, thus 1 and e solve the homogeneous problem.

1 ax ax A Special Case, (D ) Applied To e Notice in the case of exponential, we may factor out e in the formal expansion since dn eax = aneax dxn Thus 1 eax a a2 eax eax = 1+ + + ... = D 2 a 

2 Tutorial #5 – Winter 2016 MAT 267 where we side-stepped the notion of convergence once again, but clearly this is an inverse since 1 eax (D ) eax =(D ) = eax D a ✓ ◆ Since this will work with any a C and the inverse raised to integer powers, we’ve therefore found a way to 2 handle exponentials. The only issue that may occur is if a = since the expansion isn’t deﬁned (in other words, a is an eigenvalue). This can easily be ﬁxed using the exponential shift theorem,

L(D)[eaxy]=eaxL(D + a)[y] when L(x) is a polynomial (the proof goes by induction, and also applies to the inverse). Thus if is a root of L,i.e.L(D)=(D )kg(D) and g() = 0, we see that 6 1 1 xk ex = ex 1=ex (D )kg(D) Dkg(D + ) k!g() Note there is a similar version from the Laplace Transform which is deﬁned as

1 [y(t)](s) esty(t)dt L ⌘ Z0 which is another useful tool for solving ODE’s. It takes the form [eatf(t)] = [f(t a)]. L L pg. 282 - # 32 Solve

y000 + y0 = cos x

Solution Well, in terms of D we have that

D(D i)(D + i)y = cos x Now since we’ve just dealt with exponentials so far, note eix = cos x + i sin x, so lets solve

D(D i)(D + i)y = eix and take the real part. Letting g(D)=D(D + i) like the above, we see 1 x x x cos x x sin x y (x)= eix = eix = eix = i p (D i)g(D) g(i) 2 2 2 Since we just want the real part of the solution, we see the particular solution to the ODE is x cos x y (x)= p 2 Noting that the eigenvalues of the equation are =0, i, we have that ± x cos x y(x)=C + C cos x + C sin x 1 2 3 2 is the general solution.

Partial Fraction Decomposition with Di↵erential Operators As you’ve probably seen before with polynomials, you may decompose 1 c c = 1 + 2 (D + 1)(D + 2) D + 1 D + 2 Let’s see how this would apply to the previous example.

3 Tutorial #5 – Winter 2016 MAT 267 pg. 282 - # 32 Solve (using partial fractions)

y000 + y0 = cos x

Solution Using what we saw before, let’s try to decompose into pieces: 1 c c c = 1 + 2 + 3 D(D i)(D + i) D D i D + i This implies we need

c1 + c2 + c3 =0 1 1 (D i)(D + i)c + D(D + i)c + D(D i)c =1 = 8 c c =0 = c = ,c = 1 2 3 ) 2 3 ) 2 2 3 2 > < c1 =1 > Thus :> 1 1 1 1 = D(D i)(D + i) D 2(D i) 2(D + i) Now if we apply this to eix,wesee 1 eix xeix eix 3 xeix eix = + C = eix + C D(D i)(D + i) i 2 4i 4i 2 If we take the real part of this solution we see the following particular solution 3 x cos x y (x)= sin x + C p 4 2

Quiz Find a partial solution using any inverse operator method for

2x 2 y00 +3y0 +2y = 2(e + x )

Solution We see that L(D)=(D + 2)(D + 1) Thus we want to solve 1 2x 2 y (x)= (2e +2x ) p (D + 2)(D + 1) For the exponential, we may use what we’ve previous talked about to ﬁnd that we have L =(D + 2)g(D), hence

2x 2xe 2x y (x)= = 2xe pe g( 2) For the polynomial, we have that 1 1 D D2 1 3D 7D2 y (x)= 2x2 = 1 + 1 D + D2 2x2 = 1 + 2x2 pp (D + 2)(D + 1) 2 2 4 2 2 4 ✓ ◆ ✓ ◆ Thus we see 3 y (x)=x2 3x + pp 2 So 2x 2 7 y (x)= 2xe + x 3x + p 2 is a particular solution.