Solving a nonlinear, first order ODE with a trig function

Question

I have an equation of the form,

$$ \frac{dy}{dx} = A(x) + B(x) \sin^2(y)$$

How would I go about solving this? Obviously were $A(x) = 0$ then this would be separable and all would be OK, however the presence of the $A(x)$ term is confusing me.

Thanks in advance for any guidance

Answer 1

Divide both sides by $\sin^2(y)$ to get

$$\csc^2(y)\frac{\mathrm dy}{\mathrm dx}=A(x)\csc^2(y)+B(x)$$

Apply a Pythagorean identity:

$$\csc^2(y)\frac{\mathrm dy}{\mathrm dx}=A(x)\cot^2(y)+A(x)+B(x)$$

Let $u=\cot(y)$ to get

$$-\frac{\mathrm du}{\mathrm dx}=A(x)u^2+A(x)+B(x)$$

which is a Riccati equation.

Answer 2

You shouldn't expect a closed-form solution for general $A(x)$ and $B(x)$. Even for $A(x)=x$ and $B(x)=1$, or $A(x)=1$ and $B(x)=x$, Maple finds no closed-form solution. Of course numerical or series solutions are available.

Answer 3

You can solve it when $A(x)=B(x)$, then $$\frac{dy}{1+\sin^2 y}=A(x) dx \implies \int\frac{dy}{1+\sin^2y}=\int A(x) dx+C$$ $$\implies \int \frac{du}{1+2u^2}=\int A(x) dx +C \implies \frac{1}{\sqrt{2}} \tan^{-1} u\sqrt{2}, u=\tan y.$$ $$y(x)=\tan^{-1}\left\{\frac{1}{\sqrt{2}}\tan\left[ \sqrt{2}\left( \int A(x) dx+C \right)\right]\right\}$$