Second Order ODE's with variable cofficients

Question

I am having trouble solving the following ODE $$x^2\ddot{y}+2x\dot{y}-2y=0$$ The only other pieces of information I am given are that $y(1)=0$ and $y(2)=\frac{7}{2}$. I would attempt to use a variation of parameters but I do not have a known $y_1(t)$. Is it possible to solve this or am I missing a vital piece of information? Any help would be great!

Answer 1

You can try reduction of order since $y=x$ is a solution. Then try $y(x)={v(x)}{x}$

Or you can use this method..

$$x^2y''+2xy'-2y=0$$

Substitute $y=x^m$ $$m^2+m-2=0 \implies m_1=1,m_2=-2$$ then the solution is $$y= {c_1}x^{m_1}+c_2x^{m_2}$$ $$y= {c_1}x+\frac {c_2}{x^2}$$

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You can also integrate directly $$x^2y''+2xy'-2y=0$$ $$y''+2\frac {(xy'-y)}{x^2}=0$$ $$y''+2(\frac yx)'=0$$ $$y'+\frac {2y}x=K_1$$ Which is of first order..