# Second-Order Nonlinear Differential Equation

• MHB
• sav26
In summary, the conversation discusses a non-linear differential equation and its solution in terms of circular motion. The equation is similar to the equation of motion of an object under Newtonian gravity and has a solution in the form of a conic section with the origin as a focal point. The general solution of the equation is y(t)=r(t)(cosθ(t), sinθ(t)), and the specific solution is the unit circle, y(t)=(cos t, sin t).

#### sav26

$$\displaystyle \begin{cases} y''(t)=-\frac{y(t)}{||y(t)||^3} \ , \forall t >0 \\ y(0)= \Big(\begin{matrix} 1\\0\end{matrix} \Big) \ \text{and} \ y'(0)= \Big(\begin{matrix} 0\\1\end{matrix} \Big) \end{cases} \\ y(t) \in \mathbb{R}^2 \ \forall t$$

Non-linear differential equations are, in general, extremely difficult and most simply do not have solutions in terms of elementary functions. Do you have any reason to believe this does or will a numerical solution suffice?

it must have solutions yes, it's in my homework and the following question requires these solutions

If I can see and remember correctly, this equation is similar than equation of motion of an object under Newtonian gravity. Thus the solution indeed exists, and in general the shape of the solution $$y(x)$$ can be derived, but the time depense of coordinates $$(x(t), y(t))$$ can be impossible to write down. However, everything is easier in circular motion... :unsure:

Theia said:
If I can see and remember correctly, this equation is similar than equation of motion of an object under Newtonian gravity. Thus the solution indeed exists, and in general the shape of the solution $$y(x)$$ can be derived, but the time depense of coordinates $$(x(t), y(t))$$ can be impossible to write down. However, everything is easier in circular motion...
Yeah, it is indeed the motion of an object in a field of central gravity.
So the solution is a conic section (ellipse, hyperbola, or parabola) with the origin as a focal point.
That is, the general solution of the differential equation is
$$y(t)=r(t)(\cos\theta(t), \sin\theta(t))$$
with $r(t)=\frac{b^2}{a-c\cos\theta(t)}$ and $r(t)^2 \theta'(t) = \text{constant}$.

Since every constant is $0$ or $1$, we can see by inspection that the solution is the unit circle.
That is
$$y(t) = (\cos t, \sin t).$$
Things are indeed easier in circular motion. :geek:

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