y′′ +(3/4-x)y' +(2/x-4)^2 y=X

Question

Solve this eular ordinary difffrentiual equation

y′′ +(3/4-x)y' +(2/x-4)^2 y=X

Answer 1

Let X=x-4:

y"-(3/X)y'+(2/X)²y=X+4 if given "X" should be "x".

Multiply through by X²:

X²y"-3Xy'+4y=X³+4X².

If one of the solutions to X²y"-3Xy'+4y=0 is y=aX^r, (a is a non-zero constant) then:

since y'=dy/dx=dy/dX=arX^r-1, y"=d²y/dx²=d²y/dX²=ar(r-1)X^r-2 and:

ar(r-1)X^r-3arX^r+4aX^r=0=aX^r(r(r-1)-3r+4)=aX^r(r-2)², r=2 because X^r≠0 and a≠0.

[NOTE: r(r-1)-3r+4 is the expression arising from Euler's Method. Here it has been derived from scratch.]

y=aX²=a(x-4)² is a solution of X²y"-3Xy'+4y=0 (that is, y"+(3/(4-x))y'+(2/(x-4))²y=0).

If the original DE was intended to be: y"+(3/(4-x))y'+(2/(x-4))²y=0 then Euler's Method is in fact essentially the solution above.

However, if the DE was intended to be: y"+(3/(4-x))y'+(2/(x-4))²y=x, we still need a solution which involves more than Euler's Method.

If y=y₁+y₂ is the full solution and y₁=a(x-4)², then y₂ (to be found) is a solution to the original DE.

If, though, we start with y"+(3/(4-x))y'+(2/(x-4))²y=X where X=x-4 (not given, but assumed), then let y=aX²+bX³:

y'=2aX+3bX², y"=2a+6bX

y"-(3/X)y'+(2/X)²y=X, multiply through by X²:

X²y"-3Xy'+4y=X³,

2aX²+6bX³-3X(2aX+3bX²)+4(aX²+bX³)=X³,

2aX²+6bX³-6aX²-9bX³+4aX²+4bX³=X³,

bX³=X³⇒b=1, so y=aX²+X³=X²(a+X)=(x-4)²(a+x-4), making the solution:

y=(x-4)²(x+a-4) where a≠0 is a constant.