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.

Solve the differential equation $$xy^3y'=2y^4+x^4$$

share|improve this question
add comment

3 Answers

up vote 8 down vote accepted

Hint: Introduce the new function $z(x)$ with $y(x)= x z(x)$ to make the ODE separable.

Ater some calculation, you should obtain $$ \frac{ z^3z'}{1+z^4} = \frac1x.$$

share|improve this answer
    
Ah, got it! Very helpful! Thanks! (+1) –  Chris's sis Jan 2 '13 at 10:49
add comment

Dividing $xy^3$ on both sides, we get $$\tag{1}y'=2\frac{y}{x}+\frac{x^3}{y^3}$$ which is homogenous differential equation. We can use the substitution $u=\frac{y}{x}$. Then $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$. Substituting it into $(1)$, we obtain $$x\frac{du}{dx}+u=2u+\frac{1}{u^3}$$ or $$x\frac{du}{dx}=u+\frac{1}{u^3}$$ which can be solved by separation of variables.

share|improve this answer
    
nice way to choose (+1) –  Chris's sis Jan 2 '13 at 10:50
add comment

Another method is to realise that $y^3y'$ is, up to a constant factor, the derivative of $y^4$. So we obtain:

\begin{align*} xy^3y' &= 2y^4 + x^4\\ x\frac{d}{dx}\left(\frac{y^4}{4}\right) - 2y^4 &= x^4\\ \frac{d}{dx}(y^4) - \frac{8}{x}y^4 &= 4x^3\\ \end{align*}

where, in the last step, we need $x \neq 0$. Now you can solve this by using an integrating factor.

share|improve this answer
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.