Nonlinear second order differential equation involving sine

Question

I need to solve this ODE:

$A +B\frac{dy}{dt}+C\sin(y)+\frac{d^2y}{dt^2} = 0$

where A, B, and C are constants, and $y$ is $y(t)$.

I manage to reduce the order of the ODE by using dimension analysis to:

$A +B\frac{dy}{dt}+\sin(y) = 0$

where A and B are constant, and $y$ is $y(t)$

I would like to know the analytic solution for the 1st order ode (and if it's a possibility for the 2nd order as well).

I'm working on a numerical solution just in case there is no solution for that problem, in the meanwhile.

Thanks in advance!

Answer 1

$$y''(x)+By'+c \sin y +A=0~~~~(1)$$ Let $y'=p$, then $$\frac{dp}{dx}+Bp+c \sin y+A=0~~~~(2)$$ $\frac{dp}{dx}=p\frac{dp}{dy}$, then $$p\frac{dp}{dy}+Bp+C \sin y+A=0$$ $$pdp+(Bp+C \sin y+A)dy=0~~~(3)$$ Multiply (3) by $e^{By}$ $$p~e^{By} dp+e^{By}(Bp+C \sin y+A)dy=0~~~(3)$$ (3) becomes an Exact Eq. Whose solution is $$\int pe^{By} dp \text{(treat $y$ as const)}+\int e^{By}(C \sin y+A) dy=G$$ $$\implies e^{By}\frac{p^2}{2}+e^{By}\left( \frac{B^2C \sin y- BC \cos y+A+AB^2}{B+B^3}\right)$$ $$\implies p=\frac{dy}{dx}=\sqrt{\sqrt{2}G+\sqrt{2}\frac{Bc\sin y-B^2C \sin y-A-AB^2}{(B+B^3)}}$$ Given the values of A,B,C, this integral can be solved to get $y(x)$ with two constants $G$ and one more $H$.

Answer 2

$$y''+by'+c\sin y+a=0$$ Multiply it by $2y'$ to get $$2y'y''+2by'^2+2c\sin y y'+2ay'=0$$ $$\implies \frac{dy'^2}{dx}+2by'^2=-(2c\sin y+2a) \frac{dy}{dx}.$$ $$\frac{d (e^{2by}y'^2)}{dx}=-2(c\sin y+2) \frac{dy}{dx} e^{2by}$$ Integrate both sides w.r. $x$, to get $$y'^2e^{2by}=-2\int e^{2by}(c\sin y+a) dy$$ $$y'^2=-2e^{-2by}\int e^{2by}(c\sin y+a) dy +ge^{-2by}$$ From here one can carry on.

Answer 3

Considering$$A +B\,y'+C\sin(y)+y'' = 0$$ start switching variables to make $$A+\frac B {t'}+C\sin(y)-\frac{t''}{[t']^3}=0$$ which makes the reduction of order quite obvious. So $$A+\frac B {p}+C\sin(y)-\frac{p'}{p^3}=0$$ If I am not mistaken, we should arrive to something like $$\frac 1 p=-\frac A B+\frac{C (\cos (y)-B \sin (y))}{B^2+1}+k\, e^{-B y}$$

Doing the same with

$$A +B\,y'+\sin(y) = 0\implies t'=-\frac{B}{A+\sin (y)}$$ which is quite simple using the tangent half-angle subsitution.

Making the story short, we should get $$t=k-\frac{2 B }{\sqrt{A^2-1}}\tan ^{-1}\left(\frac{A \tan \left(\frac{y}{2}\right)+1}{\sqrt{A^2-1}}\right)$$ which can be inversed.