Equations of Trajectories of : $x'=1, y' = 2(1-x)\sin(1-x^2)$

Question

Studying for my Dynamical Systems course, I came upon an exercise that I do not know how to proceed at :

Exercise :

Find the Equations of the Trajectories of the system : $x'=1, y' = 2(1-x)\sin(1-x^2)$

Discussion :

I've handled simpler systems, which have closed form solution and you can derive the trajectory equations, for example like, when a solution is $(\cos t, \sin t)$ the trajectory equation is the unit circle.

But in this particular exercise, despite that $x'=1$ has a simple and easy solution, the second differential equation cannot be solved in a known-function form way, as Wolfram Alpha says, which means that I do not know what to do in order to find the equations of the trajectories.

(Neither the substitution $\frac{dy}{dy} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ works because it's essentially the same differential equation)

I would really much appreciate some thorough help !

Answer 1

From the first equation, you see that

$$x(t)=t+c_1$$

Using this for the second equation we obtain

$$y'(t)=2(1-t-c_1)\sin\left[1-(t+c_1)^2\right]$$

Integration is possible by introducing the Fresnel cosine and sine integrals

$$C(u)=\int_{0}^{u}\cos\left[\dfrac{\pi}{2}\tau^2\right]\tau^2d\tau$$ $$S(u)=\int_{0}^{u}\sin\left[\dfrac{\pi}{2}\tau^2\right]\tau^2d\tau.$$

Using these definitions we obtain

$$y(t)= -\sqrt {2\pi} \left[ \cos 1 \,S \left( {\frac {\sqrt {2} \left( t+c_1 \right) }{\sqrt {\pi }}} \right) -\sin 1 \, C \left( {\frac {\sqrt {2 } \left( t+c_1 \right) }{\sqrt {\pi }}} \right) \right] $$ $$-\cos \left( {c_1}^{2}+2\,tc_1+{t}^{2}-1 \right) +c_2.$$

Now resubstitute $t=x-c_1$ into $y(t)$ to eliminate time and to obtain $y(x)$ as

$$y(x)= -\sqrt {2\pi} \left[ \cos 1 \,S \left(\sqrt{2/\pi}x \right) -\sin 1 \, C \left( \sqrt{2/\pi}x \right) \right] $$ $$-\cos \left( {c_1}^{2}+2\,(x-c_1)c_1+(x-c_1)^{2}-1 \right)+c_2$$

$$= -\sqrt {2\pi} \left[ \cos 1 \,S \left(\sqrt{2/\pi}x \right) -\sin 1 \, C \left( \sqrt{2/\pi}x \right) \right]$$ $$-\cos \left(x^2-1\right)+c_2$$

I don't think that there is a way to express this result by using only elementary functions.

Answer 2

this is right what you have $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ and this $$\frac{dy}{dx}=2(1-x)\sin(1-x^2)$$ solve this equation.