Solve Differential Equation 2: $\frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+2y=0$

Question

solve $\frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+2y=0$ I tried to solve through linear differential equation but it has variable with x in first derivative .

Answer 1

Substituting $$y=xv$$ then we get $$y'=xv'+v$$ and $$y''=xv''+2v'$$ then our equation will be

$$xv''-2v'(x^2-1)=0$$ now we Substitute

$$v'=u$$ then we get $$\frac{u'}{u}=-\frac{2}{x}+2x $$

and we get

$$\int\frac{u'}{u}dx=\int\left(-\frac{2}{x}+2x\right)dx$$

this gives

$$\log(u(x))=x^2-\log(x)+C$$ Can you finish?

Answer 2

$y=c_1x$ is a solution.

You can try to integrate the differential by reduction of order

Try

$$y_1=v(x)x$$ The original equation $$y''-2xy'+2y=0$$ becomes $$v''x+2v'(1-x^2)=0$$ Substitute $w=v'$ $$w'x+2w(1-x^2)=0$$

It's a first order,differential equation $$(\ln w)'=2x-\frac 2x$$ $$\ln w=x^2-2\ln x + K$$ $$w=C_1\frac {e^{x^2}}{x^2}$$ $$v'=C_1\frac {e^{x^2}}{x^2}$$ $$v(x)=C_1\int \frac {e^{x^2}}{x^2}dx+C_2$$ $$y(x)=C_1x\int \frac {e^{x^2}}{x^2}dx+C_2x$$ $$\boxed{y(x)=C_1(x\sqrt {\pi}\text {erfi(x)}-e^{x^2})+C_2x}$$