0
$\begingroup$

Solve $y'+y=xy^3$

So I divided by $y^3$ on both sides getting $\frac{\text{dy}}{\text {dx}y^{3}}+y^{-2}=x$

Substituted $u=y^{-2}$, $\frac{\text{du}}{\text{dx}}=-2y^{-3}\frac{\text{dy}}{\text{dx}}$

Got the linear equation $\frac{\text{du}}{\text{dx}}-2u=-2x$

Found the integrating factor of $e^{-2x}$

Multiplied both sides by the integrating factor to get $e^{-2x}\frac{\text{du}}{\text{dx}}-2e^{-2x}u=-2xe^{-2x}$

Integrated both sides to get $ue^{-2x}=\int-2xe^{-2x}dx$ The right hand side I got from basically the product rule. The left hand side tured out to be $4xe^{-2x}+8e^{-2x}+C$ from tabular method of integration

So we have $ue^{-2x}=4xe^{-2x}+8e^{-2x}+C$

After simplification (I cancelled out the $e^{-2x}$, don't know if that's allowed):

I got the solution of $1=4xy^2+8y^2+C$, however the correct answer is $\frac{1}{y^2}=Ce^{2x}+x+\frac{1}{2}$

Where did I go wrong?

$\endgroup$
  • 1
    $\begingroup$ First, your integration is a bit off. Second, the canceling out of $e^{-2x}$ is multiplication by $e^{2x}$. $\endgroup$ – Lev Borisov Mar 24 '16 at 21:10
  • $\begingroup$ Yeah I think I used the tabular method incorrectly $\endgroup$ – shoestringfries Mar 24 '16 at 21:11
2
$\begingroup$

After finding the integrating factor, you should have

$$\begin{align}\frac{d(e^{-2x} u)}{dx} = -2xe^{-2x} &\implies e^{-2x} u = \color{red}{\frac{1}{2} e^{-2x}(2x + 1)} + C \\ &\implies u = \frac{1}{2} (2x+ 1) + Ce^{2x} \end{align}$$

where on the second implication you multiply both sides by $e^{2x}$.

This gives you the result.

$\endgroup$
  • $\begingroup$ So the constant has to indicate you multiplied by $e^{2x}$ because there is a variable $x$ included, right? $\endgroup$ – shoestringfries Mar 24 '16 at 21:19
  • $\begingroup$ I'm not sure what you mean by "indicate", but, yes, there is a variable $x$ and the family of solutions is given by $$\frac{1}{y^2} = Ce^{2x} + x + \frac{1}{2}$$ and the integration in red is by parts. $\endgroup$ – Aaron Maroja Mar 24 '16 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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