Question

Use change of variables to find general solution of this equation

Answer 1

use \[y=x^{\frac{ 1 }{ 2 }}v(x)\] to find general solution of the equation \[x^2y''+2xy'+ \lambda ^2 x^2y = 0,. x>0\]

Answer 2

\[\begin{align*}y&=x^\frac{1}{2}~v(x)\\
y'&=\frac{1}{2}x^{-\frac{1}{2}}~v(x)+x^\frac{1}{2}~v'(x)\\
y''&=-\frac{1}{4}x^{-\frac{3}{2}}~v(x) + \frac{1}{2}x^{-\frac{1}{2}}~v'(x)+\frac{1}{2}x^{-\frac{1}{2}}~v'(x)+x^\frac{1}{2}~v''(x)\\
&=x^\frac{1}{2}~v''(x)+x^{-\frac{1}{2}}~v'(x)-\frac{1}{4}x^{-\frac{3}{2}}~v(x)
\end{align*}\]
\[x^2y''+2xy'+\lambda^2x^2y=0\]
Substituting, you get
\[x^2\left[x^\frac{1}{2}~v''(x)+x^{-\frac{1}{2}}~v'(x)-\frac{1}{4}x^{-\frac{3}{2}}~v(x)\right]+2x\left[\frac{1}{2}x^{-\frac{1}{2}}~v(x)+x^\frac{1}{2}~v(x)\right]\\
~~~~~~~~~~~~~~+\lambda^2x^2\left[x^\frac{1}{2}~v(x)\right]=0\\
x^\frac{5}{2}~v''(x)+x^{\frac{3}{2}}~v'(x)-\frac{1}{4}x^{\frac{1}{2}}~v(x)+x^{\frac{1}{2}}~v(x)+2x^\frac{3}{2}~v(x)+\lambda^2x^\frac{5}{2}~v(x)=0
\]
Since \(x>0\), you can divide both sides by \(x^\frac{1}{2}\), so you have
\[\begin{align*}x^2~v''(x)+x~v'(x)+\frac{3}{4}v(x)+2x~v(x)+\lambda^2x^2~v(x)&=0\\
x^2~v''(x)+x~v'(x)+\left(\lambda^2x^2+2x+\frac{3}{4}\right)~v(x)&=0\end{align*}\]
Unfortunately, I don't know where to go from here...

Answer 3

i thnk i can try and and do it from here. this helps a lot. i really thank uuuu have a amazing day

Answer 4

i will try to go from here now if i can

Answer 5

Look's like @SithsAndGiggles is on the right path. If you notice now with his result, you have non-euler equation so you can use power series to solve this. This problem will take a really long time so when you're ready you we can work it step by step. Since x = 0 is a regular singular point I think you will be guessing that \(\huge\sum_{n=0}^{\infty}c_nx^{n+r}\)

Answer 6

ty!!!!!!!!!!!!!!!!!!! i thank u so much!!