How do I solve the differential equation $x^2y'' + 2xy' = 0$, where $y(1) = 1$ and $y'(1) = 1$. For some reason I'm running into trouble, any tips or hints greatly appreciated.
I started by finding the characteristic equation for a Cauchy Euler diff eq: $ar²+(-a+b)r+c=0$. Where do I go from here?
On
$x^2y'' + 2xy' = 0\tag{1}$
Yes. It is in Cauchy-Euler form.
Substitute $t=\log x$
Then transformed equation is given by $$\frac{d^2y}{dt^2}+(2-1) \frac{dy}{dt}=0\tag{1}$$
Auxiliary equation of $2$ : $r^2+(2-1) r+0=0$
implies $r=0, -1$
Solution of $(2)$ : $y(t)=C_1+C_2e^{-t}$
Solution of $(1) $ : $y(x) =C_1+\frac{C_2}{x}$
Initial conditions:
$1=y(1) =C_1+C_2\tag{3}$
$1=y'(1) =-C_2\tag{4}$
Solving $(3), (4) $ we have $C_1=2, C_2=-1$
Solution of $(1) $ : $y(x) =2-\frac{1}{x}$
The ordinary differential equation $$x^2y'' + 2xy' = 0$$ can be easily recast into $$(x^2 y')' = 0$$ in view of the product rule for differentiation. It follows that $$x^2y' = c_1$$ where $c_1 \in \Bbb R$, and then it remains to solve $$y' = \frac{c_1}{x^2}$$ using direct integration to get $$y = -\frac{c_1}{x} + c_2$$ where $c_2\in \Bbb R$. To find $c_1$ and $c_2$, use the initial conditions $y(1) = 1$ and $y'(1) = 1$. I leave the computation to you - we get $c_1 = 1$ and $c_2 = 2$.