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Second-Order Nonlinear Differential Equation

sav26

New member
Nov 21, 2020
2
Hi there can someone please help me with this differential equation, I'm having trouble solving it
\(\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
\)

Thanks in advance ^^
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
756
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?
 

sav26

New member
Nov 21, 2020
2
it must have solutions yes, it's in my homework and the following question requires these solutions
 

Theia

Well-known member
Mar 30, 2016
115
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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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