# Sketching Solution Curves for $y'=y^{4}+1$

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.

• 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. – Galc127 Oct 27 '16 at 9:41
• 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. – H. H. Rugh Oct 27 '16 at 9:41

## 2 Answers

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.

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