15.4 EXERCISES
15.23 Prove that the general solution of
(x−2)d^2 y
dx^2+3
dy
dx+
4 y
x^2=0
is given byy(x)=1
(x−2)^2[
k(
2
3 x−
1
2
)
+cx^2]
.
15.24 Use the method of variation of parameters to find the general solutions of
(a)d^2 y
dx^2−y=xn,(b)d^2 y
dx^2− 2
dy
dx+y=2xex.15.25 Use the intermediate result of exercise 15.24(a) to find the Green’s function that
satisfies
d^2 G(x, ξ)
dx^2
−G(x, ξ)=δ(x−ξ)withG(0,ξ)=G(1,ξ)=0.15.26 Consider the equation
F(x, y)=x(x+1)d^2 y
dx^2+(2−x^2 )dy
dx−(2 +x)y=0.(a) Given thaty 1 (x)=1/xis one of its solutions, find a second linearly inde-
pendent one,
(i) by settingy 2 (x)=y 1 (x)u(x), and
(ii) by noting the sum of the coefficients in the equation.
(b) Hence, using the variation of parameters method, find the general solution
of
F(x, y)=(x+1)^2.15.27 Show generally that ify 1 (x)andy 2 (x) are linearly independent solutions of
d^2 y
dx^2+p(x)dy
dx+q(x)y=0,withy 1 (0) = 0 andy 2 (1) = 0, then the Green’s functionG(x, ξ) for the interval
0 ≤x, ξ≤1andwithG(0,ξ)=G(1,ξ)=0canbewrittenintheformG(x, ξ)={
y 1 (x)y 2 (ξ)/W(ξ)0<x<ξ,
y 2 (x)y 1 (ξ)/W(ξ) ξ<x< 1 ,whereW(x)=W[y 1 (x),y 2 (x)] is the Wronskian ofy 1 (x)andy 2 (x).
15.28 Use the result of the previous exercise to find the Green’s functionG(x, ξ)that
satisfies
d^2 G
dx^2
+3
dG
dx+2G=δ(x−x),in the interval 0≤x, ξ≤1, withG(0,ξ)=G(1,ξ) = 0. Hence obtain integral
expressions for the solution ofd^2 y
dx^2+3
dy
dx+2y={
00 <x<x 0 ,
1 x 0 <x< 1 ,distinguishing between the cases (a)x<x 0 ,and(b)x>x 0.