# Solving 2nd order ODE with conditions (problem)

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. Im just wondering, could you explain how you got from the 2nd line (right arrow) to the 3rd line (right arrow)? I am not sure how you got $y_1d(1+x)$ in your 2nd line, i see you have multiplied by $dx$ from the 1st line to the 2nd line, but how did the $y_1d(1+x)$ come about? Thanks :) –  Heijden Mar 16 '12 at 2:41
You are welcome Hejden! Here $d$ means derivative, so e.g., $d(x^2)=2x$ and so on. Now note that $d(1+x)=d(1)+d(x)=0+dx=dx$ and finally, $d(uv)=u.dv+v.du$. –  Tapu Mar 16 '12 at 8:54
sorry $d(x^2)=2x.dx$. –  Tapu Mar 17 '12 at 7:23
thanks Tapu, now I know how to solve it :) –  Heijden Mar 17 '12 at 17:58