Question

How to integrate a non-linear differential equation?

Answer 1

Here is a way to make it into a first degree equation. We have

y′′(x)+axcosy(x)=0

So we can multiply by y′(x) to get

y′′y′+axcos(y)y′=0

Now we can integrate to get

∫y′′y′dx+∫axcos(y)y′dx=∫0dx

For the first integral, we can set u=y′, du=y′′dx to get

∫uy′′dx=∫udu=u2+C=y′(x)2+C1

For the second rule we use product rule with f(x)=ax and g′(x)=cos(y)y′

So we have f′(x)=a and g(x)=sin(y)

Now we apply integration by parts to obtain

∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx=f(x)g(x)−∫ag′(x)dx

=f(x)g(x)−ag(x)+C2=a(x−1)sin(y)+C2

Then we have your first order differential equation

y′(x)2+a(x−1)sin(y(x))=C