Solve by using variation of parameters

[SOS2012]

x²y"(x)-3xy'(x)+3y(x)=2(x^4)(e^x)

=>y"(x)-(3/x)y'(x)+(3/x²)y(x)=2x²e^x

i don't know how to approach this problem because the coefficients are not constant and i am used to being given y1 and y2

HELP!

 

[HallsofIvy]

This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t.

$$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$
and, differentiating again,
$$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$

 

[SOS2012]

This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t.

$$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$
and, differentiating again,
$$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$

thank you very much. i appreciate the help