Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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,....

share|improve this answer
    
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
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.