Solve x^2 times second derivative -3x times first derivative -5y =sin(logx) my solution is integrating both sides we have sinxlogx

Answer 1

Your school subject is "St. Eugene university"? That must be very interesting.

Answer 2

First of all: St Eugene university is not a school subject

second: I can't make any sense out of your post.

Is it:
x^2 * (second derivative of -3x) * (first derivative of -5y) = sin(logx) ??

is it the derivative with respect to x ?
then 2nd derivative of -3x is 0, and
you have
x^2 * (0) * (first derivative of -5y) = sin(logx)
0 = sin(logx)
then logx = 0, not possible
or logx = π , x = appr 1385.46
(check: log 1385.46 = 3.14159... )
or logx = 2π , x = 1,919,487.58..
or logx = 3π , x = getting silly

Answer 3

I interpret the somewhat garbled language as

x^2 y" - 3xy' - 5y = sin(logx)

well, the homogeneous equation has solution

y = ax^5 + b/x

Now apply the methods for the Euler equation to get the desired solution.