Question

By changing the independent variable.solve the differential equation. d2x/dx² - (dy/dx)(1/x) + 4x²y = x⁴

Answer 1

Let $t=x^2,x=\sqrt{t}$

$y''-y'/(\sqrt{t})+4yt=t^2$

$\frac{dy}{dx}=2\sqrt{t}\frac{dy}{dt},\frac{d^2y}{dx^2}=4t\frac{d^2y}{dt^2}+2\frac{dy}{dt}$

$4t\frac{d^2y}{dt^2}+4ty(t)=t^2$

$\frac{d^2y}{dt^2}+y(t)=t/4$

$k^2+1=0$

$k=\pm i$

The general solution:

$y(t)=c_1cost+c_2sint$

For the particular solution:

$\tilde{y}(t)=A+Bt$

Then:

$A+Bt=t/4$

$A=0,B=1/4$

$\tilde{y}(t)=t/4$

So:

$y(t)=t/4+c_1cost+c_2sint$

Answer:

$y(x)=x^2/4+c_1cos(x^2)+c_2sin(x^2)$