I ran into the following problem while doing revision:

Sketch a solution curve of the differential equation $y' = y^{4}+1$

I tried to solve the equation using exact equations and integrating factors, although I didn't get anywhere. Wolfram Alpha gives out a very convoluted answer, so I was wondering if there was a simpler way to do it. I think for the easier case $y' = 1+y^{2}$ one would get as a solution curve $y=\tan(x)$, but not sure on how to go about this one.

  • 1
    $\begingroup$ Hint: $$\frac{1}{y^4+1}=\frac{1}{y^4+2y^2+1-2y^2}=\frac{1}{(y^2+1)^2-(\sqrt{2}y)^2}=\frac{1}{(y^2-\sqrt{2}y+1)(y^2+\sqrt{2}y+1)}$$ Now you can use partial fractions and integrate it. Edit: missed the part that you need to sketch it, not solving the equation. $\endgroup$ – Galc127 Oct 27 '16 at 9:41
  • 5
    $\begingroup$ You shouldn't try to seek an explicit solution. Instead draw the vector field $(t,y)\mapsto (1,1+y^4)$ in the $(t,y)$ plane and find integral curves tangent to the vector field. $\endgroup$ – H. H. Rugh Oct 27 '16 at 9:41

For small $y$ your equation is close to $y'=1$, for large $y$ close to $y'=y^4$. Both are easy to solve. After that you only need to suitably connect those approximate solutions.


$$\frac{dy}{1+y^4}=dx$$ $$\frac{dy}{(y^2+1)^2-2y^2}=dx$$ $$\frac{dy}{(y^2+1+\sqrt{2}y)(y^2+1-\sqrt{2}y)}=dx$$ $$\frac{(y+\sqrt{2})dy}{2\sqrt{2}(y^2+\sqrt{2}y+1)}-\frac{(y-\sqrt{2})dy}{2\sqrt{2}(y^2-\sqrt{2}y+1)}=dx$$


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.