y' + y/x = x^2

can u use lpaplace transform for this?

seein how the variable is x and not t?

y' + y/x = x^2

is

y' + y/t = t^2

But why would you use a Laplace transformation here ?

You can easily solve this linear ordinary differential equation, with basic means.

cuz prof wants us to use lapalce transofrm

Laplace.

prof will be pleased

y' + y/x = x^2
L{ }: sY(s) + Y(s) / (1/s^2) = 2/s^3 use table + algebra to get transform
sY(s) + s^2 Y(s) = 2/s^3
Y(s) (1+s) = 2/s^4
Y(s) = 2 / {(s^4)(1+s)}

partial fractions decomposition

Y(s) = 2 / {(s^4)(1+s)} = A/s + B/s^2 + C/s^3 + D/s^4 + E/(s+1)
A=-2, B=2, C=-2, D=2, E=2

Y(s) = 2{ -1/s + 1/s^2 - 1/s^3 + 1/s^4 + 1/(s+1) }
L-1{ }: y(x) = -2 + 2x - x^2 + x^3/3 + 2e^(-x) inverse transform & answer