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I was reading my notes and I am given 2 equations:

$(1+x)\frac{dy_0}{dx} + y_0 =0,\ \ \ \ y_0(1) =1$

$(1+x)\frac{dy_1}{dx} + y_1 = - \frac{d^2y_{0}}{dx^2} ,\ \ \ \ y_1(1) =0$

Which have the solutions:

$y_0(x) = \frac{2}{1+x}$

$y_1(x) = \frac{2}{(1+x)^3} - \frac{1}{2(1+x)}$

The problem I have is that I can't seem to get the 2 solutions. I tried solving the equations, starting with the first one, and what I got was:

$-\ln y_0 = \ln(1+x) + c$

$y_0 = -A(1+x)$ and using $y_0(1) =1$, I get $A = -\frac{1}{2}$ and so $y_0(x) = \frac{1+x}{2}$. Have I done something wrong here?

And how do I solve $(1+x)\frac{dy_1}{dx} + y_1 = - \frac{d^2y_{0}}{dx^2} ,\ \ \ \ y_1(1) =0?$ Im kind of confused with the $y_0$ and $y_1$ terms and how to approach it.

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You have done the integration part correct, but then $-\ln y_0=\ln (1+x)+c\Rightarrow \ln y_0+\ln(1+x)=-c=A\Rightarrow y_0=\frac{A}{1+x}$

For the second part, we have $\frac{dy_0}{dx}=-\frac{2}{(1+x)^2}$. So, your equation becomes, $(1+x)\frac{dy_1}{dx}+y_1=\frac{d}{dx}\left[\frac{2}{(1+x)^2}\right]$

$\Rightarrow (1+x)dy_1+y_1d(1+x)=d\left[\frac{2}{(1+x)^2}\right]$

$\Rightarrow d\left[y_1(1+x)\right]=d\left[\frac{2}{(1+x)^2}\right]$ Integrating, $y_1(1+x)=\frac{2}{(1+x)^2}+c$ Using $y_1(1)=0$, we get $c=-2$. Hence,....

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    thanks Tapu, now I know how to solve it :)2012-03-17