0
$\begingroup$

It has been many years since I have taken diff. eq., a friend of mine is in the course and asked for my help on this problem and unfortunately it has stumped me

$(x^2 + y^4)dx = -4xy^3dy$

Any help would be great, like I said it has been a long time since I took this course but I don't seem to recall spending much time on non linear equations, so is there a standard approach for solving these types of problems?

$\endgroup$
2
  • $\begingroup$ This is an exact differential equation. $\endgroup$ Jan 12, 2017 at 2:02
  • $\begingroup$ Ah yes, it makes sense now thank you, had totally forgotten about this haha $\endgroup$
    – Elliot
    Jan 12, 2017 at 2:16

1 Answer 1

2
$\begingroup$

Building off the comment, this is an exact equation.

Let $M=x^2+y^4$ and let $N=4xy^3$. Then $\frac{dM}{dy}=4y^3$ and $\frac{dN}{dx}=4y^3.$

Let $\Psi(x,y)=\int M\,dx=\frac{x^3}{3}+y^4x+h(y).$

Now we need to figure out what $h(y)$ is.

$\Psi_y(x,y)=4y^3x+h'(y)$

Now set this equal to $N$:

$4xy^3=4y^3x+h'(y)$

Or,

$h'(y)=0$

Integrate w.r.t $y$,

$h(y)=c$.

Putting it all together:

$\Psi(x,y)=\frac{x^3}{3}+y^4x+c$

$\endgroup$
3
  • $\begingroup$ Thank you I had just finished working that out as you posted it =) $\endgroup$
    – Elliot
    Jan 12, 2017 at 2:18
  • $\begingroup$ Thanks for the question, I'm trying to refresh my calc and diff-eq knowledge. $\endgroup$
    – emka
    Jan 12, 2017 at 2:18
  • $\begingroup$ I see the error now. I believe it should be fixed. $\endgroup$
    – emka
    Jan 12, 2017 at 20:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .