Solving this ODE

Question

I'm having a struggle solving this equation. y1 is one of the bases for solutions.

$$x\frac{d^2y}{dx^2}-(1+x)\frac{dy}{dx}+y=0$$ $$y_1=1+x$$ Can anyone show me how to solve this equation by steps? Thank you.

Answer 1

you already have one solution. you need a second one to form a basis of the space of solutions since it's a second order ODE. you can proceed in different ways. one approach is the method of variation of parameters.

suppose the 2nd solution is of the form $j(x)y_1(x)$. and plug that in into the ode.

after that you'll get :

$$x(x+1)j''(x)-(x^2+1)j'(x) = 0$$

can you take it from here ?

also after you're done with the computation you could compute the Wronskian to make sure they're independent.

Answer 2

By visual inspection one can come to the factorization $$ \left(x\frac{d}{dx}-1\right)\left(\frac{d}{dx}-1\right)y=0 $$ or in separate steps $$ y'-y=v\\ xv'-v=0 $$ where the second can be easily integrated to $\frac{v}x=C$, while the first has the general solution form $y=Ae^x+Bx+D$ with coefficient conditions $-B=C$, $B-D=0$ so that $$ y=Ae^x+B(1+x). $$

Answer 3

$$x\frac{d^2y}{dx^2}-(1+x)\frac{dy}{dx}+y=0$$ $$xy''-y'-xy'+y=0$$ $$x(y''-y')-(y'-y)=0$$ Substitute $z=y'-y$ $$xz'-z=0$$ Which is a first order differential equation easy to solve $$\int \frac {dz}{z}=\int \frac {dx} x=\ln|x|+K \implies z=Kx$$ $$y'-y=Kx$$ Take $e^{-x}$ as an integrating factor... $$(ye^{-x})'=Kxe^{-x}$$ Just integrate.. $$\boxed{y=Ke^{x}\int xe^{-x} dx}$$ $$.....$$