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 need some help, I've tried to solve it since yesterday but i failed.As usually, I need to find $y(x)$ which is the solution of the differential equation. Here is the equation:
$$ (1+x^2)y^3dx-(y^2-1)x^3dy = 0,\qquad y(1) = -1 $$

It is supposed to be easy, but I didn't find the right theorem nor formula to use.

Edit: I already developed to get the following equation $$ \frac {1+x^2}{x^3}dx = \frac {y^2-1}{y^3}dy $$
and after integration got:
$$ \frac {-1}{2x^2}+\ln(|x|) = \frac {1}{2y^2}+\ln(|y|) $$
but how can I get $y(x)$ from there ?

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

Please note that not every equations of the form $F(x,y)=0$ can be written as $y=f(x)$. So, here you should write the solution as $F(x,y)=0$.

It may be the case that you have found it difficult how to use the condition $y(1)=-1$. Assuming this is the case, I am giving a hint: where is your constant of integration?

So, let just add one "$C$" on the right hand side (of your last equation). The condition states that "$y$ is $-1$ when $x$ is $1$". So the value of $C$ is $-1$. Rearrange some terms to get the solution as $F(x,y)=0$.

share|improve this answer
    
If you really want explicit solutions, this one can be solved using the Lambert W function. –  Robert Israel Jun 24 '12 at 19:07
    
I have not noticed it. Please feel free to edit my post/& write your own answer/comment. –  Tapu Jun 24 '12 at 19:16
1  
Note, by the way, that the differential equation is singular on the line $y=-1$, corresponding to your curve $F(x,y)=0$ having a vertical tangent at $(1,-1)$. There will be two solutions for $x \ge 1$ and none for $x < 1$. –  Robert Israel Jun 24 '12 at 19:20
1  
Those solutions are, according to Maple, $$ -{{\rm e}^{1/2\, \left( {\it LambertW} \left( -{{\rm e}^{-{\frac {-1+2 \,\ln \left( x \right) {x}^{2}+2\,{x}^{2}}{{x}^{2}}}}} \right) {x}^{2 }-1+2\,\ln \left( x \right) {x}^{2}+2\,{x}^{2} \right) {x}^{-2}}}$$ and $$-{{\rm e}^{1/2\, \left( {\it LambertW} \left( -1,-{{\rm e}^{-{\frac {- 1+2\,\ln \left( x \right) {x}^{2}+2\,{x}^{2}}{{x}^{2}}}}} \right) {x} ^{2}-1+2\,\ln \left( x \right) {x}^{2}+2\,{x}^{2} \right) {x}^{-2}}} $$ –  Robert Israel Jun 24 '12 at 19:21
    
Wow! As I said, it should be simple (cause it's just an exercise from a basic chapter). If the explicit solution contains complicated functions as Lambert W function, then I guess that the answer should be on $F(x,y)=0$ form. Not really sure, but gonna accept the answer. –  mhfff32 Jun 24 '12 at 19:48
add comment

Hint: divide by $x^3y^3$ and the variables separate.

share|improve this answer
    
I forgot to mention that I already did that, I've just edited the question –  mhfff32 Jun 24 '12 at 15:49
add comment

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.