# Solve the differential equation: $\dot{x}(t)^2+x(t)=\sin(x)^2$

I am unable to solve this nonlinear differential equation: $$\dot{x}(t)^2+x(t)=\sin(x)^2$$ I tried with Maple without success. Is it possible to solve it without the use of numerical methods? Thanks.

• I suspect that a closed formula is out of reach. You can make some phase-plane analysis, however. – Siminore Dec 9 '13 at 12:21

If $x(t)$ is of class $C^2$ you can differentiate the equation to get $$2\dot{x} \ddot{x}+\dot{x}=\sin(2x) \dot{x}$$

$\dot{x}=0$ gives constant solutions $x(t)=x_0$, which by the original ODE must satisfy $x_0=\sin(x_0)^2$, thus reducing to $x_0=0$ alone.

Other solutions satisfy $$2 \ddot{x}=\sin{2x}-1$$

and this equation seems to be within the capability of Maple (an implicit solution can be obtained).

Another approach could be separating the variables: $$\dot{x}^2=\sin^2 x-x \\\dot{x}= \pm \sqrt{\sin^2{x}-x} \\ \frac{dx}{\sqrt{\sin^2 x-x}}=\pm dt \\ \int \frac{dx}{\sqrt{\sin^2 x-x}}=\pm(t-t_0)$$ Which also gives an implicit solution.

Note that $\sin^2x -x\geqslant0$ if and only if $x\leqslant0$ hence one can rewrite this as $\dot x=\varepsilon u(x)$ on every interval where $x\lt0$, with $u:\xi\mapsto\sqrt{\sin^2\xi-\xi}$ and $\varepsilon=\pm1$. Thus, assuming that $x(0)=x_0$ with $x_0\lt0$, $$\int_{x_0}^{x(t)}\frac{\mathrm d\xi}{u(\xi)}=\varepsilon t,$$ for every $t$ such that the LHS yields $x(t)\lt0$. This defines implicitely $x(t)$ for every $t$ such that $x(t)$ exists. Two cases arise. If $\varepsilon=+1$, then $x$ is increasing and $x(t)\to0$ when $t\to\tau(x(0))$, where, for every $y\lt0$, $$\tau(y)=\int_{y}^{0}\frac{\mathrm d\xi}{u(\xi)},$$ thus, $x$ is defined on $[0,\tau(x_0))$. If $\varepsilon=-1$, then $x$ is decreasing and $x(t)\to-\infty$ when $t\to+\infty$ because the integral $$\int\frac{\mathrm d\xi}{u(\xi)}$$ diverges at $-\infty$. Both cases are actually the same, since the maximal interval of definition of $x$ is $(-\infty,\tau(x_0))$ when $\varepsilon=+1$ and $(-\tau(x_0),+\infty)$ when $\varepsilon=-1$. Thus, one follows the same and unique solution, starting at some point given by $x_0$, either in the direction where the solution stops to be defined after a finite time or in the direction where it is defined forever.

Said differently, there exists a unique increasing function $\gamma:(-\infty,0)\to(-\infty,0)$ such that $\gamma(0-)=0-$ and $\gamma(-\infty)=-\infty$, and, for every $x_0\lt0$ and every $t$ such that $x(t)$ exists, $$x(t)=\gamma(\varepsilon t+\gamma^{-1}(x_0)),$$ either with $\varepsilon=+1$ or $\varepsilon=-1$. And an explicit formula for $\gamma$ is $$\int_{\gamma(t)}^0\frac{\mathrm d\xi}{u(\xi)}=-t.$$