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?

  • 0
    This is an [exact differential equation](https://en.wikipedia.org/wiki/Exact_differential_equation). – 2017-01-12
  • 0
    Ah yes, it makes sense now thank you, had totally forgotten about this haha – 2017-01-12

1 Answers 1

2

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:

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

  • 0
    Thank you I had just finished working that out as you posted it =) – 2017-01-12
  • 0
    Thanks for the question, I'm trying to refresh my calc and diff-eq knowledge. – 2017-01-12
  • 0
    I see the error now. I believe it should be fixed. – 2017-01-12