Solving nonlinear differential equation with squares

Question

I'm solving the following differential equation (here $y$ is a function of $t$ and $a$ is a constant) $$(y')^2 + a^2y^2 = 1.$$ I know that one solution is $$y(t) = \frac{1}{a}\sin(at),$$ but I'm struggling to find if there are any other (complex) solutions. Is there also a more general way to approach such differential equations, specifically since we have squares?

Answer 1

$$(y')^2 + a^2y^2 = 1$$ $$y' =\pm\sqrt {1- a^2y^2 }$$ No need for an integrating factor separate and integrate
$$\int \dfrac {dy}{\sqrt {1-a^2y^2}}=\pm \int dt$$ For $a\ne 0$. You need to treat the case $a=0$. And the case $1-a^2y^2=0 \implies y=\pm \dfrac 1a$.

Substitute $u=ay$: $$\int \dfrac {du}{\sqrt {1-u^2}}=\pm a\int dt$$ Substitute $u= \sin x$: $$\int {dx}=\pm at+C$$ $$x(t)=\pm at+C$$ $$u(t)=\sin (\pm at+C)$$ $$y(t)=\dfrac 1a\sin (\pm at+C)$$