Solution of a parametric differential equation

Click For Summary
The discussion centers on solving the parametric differential equation describing the motion of a vertical projectile in a cubic resisting medium, specifically the equation d²y/dt² = -a - k*(dy/dt)³. The original poster acknowledges the complexity of the equation, noting it belongs to a challenging class of nonlinear ordinary differential equations known as Abel's equations. A participant suggests using Mathematica to derive a solution, providing a complex expression that involves logarithmic and inverse functions. The conversation highlights the difficulty of finding a straightforward solution due to the nonlinear nature of the equation. Overall, the thread emphasizes the challenges in solving such parametric differential equations and the potential utility of computational tools.
patric44
Messages
308
Reaction score
40
Homework Statement
solution of a parametric equation related to vertical motion
Relevant Equations
D2y = -a-k*(Dy)^3
hi guys
i was trying to solve this differential equation ##\frac{d^{2}y}{dt^{2}}=-a-k*(\frac{dy}{dt})^{3}## in which it describe the motion of a vertical projectile in a cubic resisting medium , i know that this equation is separable in ##\dot{y}## but in order to solve for ##y## it becomes unsolvable in this form , so i came up with this parametric solution
Image 1.jpg

and i would like to solve this parametric differential equation for ##y## but don't know what is the approach for it
i will appreciate any help , thanks
 
Last edited:
Physics news on Phys.org
patric44 said:
...i was trying to solve this differential equation ##\frac{d^{2}y}{dt^{2}}=-a-k*(\frac{dy}{dt})## ...
Your ODE is a second-order linear equation with constant coefficients. It is rather straightforward to solve, simply observe that you can write it in the following form
$$ \big(e^{kt}y^\prime\big)^\prime = - ae^{kt}.$$
Now you simply have to integrate twice.
 
William Crawford said:
Your ODE is a second-order linear equation with constant coefficients. It is rather straightforward to solve, simply observe that you can write it in the following form
(ekty′)′=−aekt.
Now you simply have to integrate twice.
i am sorry i meant to write ##(\frac{dy}{dt})^3## . its not very simple in this case
 
patric44 said:
i am sorry i meant to write ##(\frac{dy}{dt})^3## . its not very simple in this case
No worries! It was me that read your original post in a hurry. Your differential equation in ##y^\prime## belong to a notorious difficult class of ODE's called Abel's nonlinear ODE's of the fist-kind. I haven't had the change nor time to study this class of ODE's, so I'm afraid that I can't provide you with any hint to how to selve it (if possible).
Is this homework?
 
Last edited:
just for kicks, i put the ODE into mathematica and got $$\left\{\left\{y(t)\to \frac{2 \log \left(\sqrt[3]{a}+\sqrt[3]{k} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]\right)-\log \left(-\sqrt[3]{a} \sqrt[3]{k} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]+k^{2/3} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]{}^2+a^{2/3}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \sqrt[3]{k} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 \sqrt[3]{a} k^{2/3}}+c_2\right\}\right\}$$
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...