Solution of a parametric differential equation

In summary: Is this homework?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
  • #1
patric44
308
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
 
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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
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  • #5
just for kicks, i put the ODE into mathematica and got $$\left\{\left\{y(t)\to \frac{2 \log \left(\sqrt[3]{a}+\sqrt[3]{k} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]\right)-\log \left(-\sqrt[3]{a} \sqrt[3]{k} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]+k^{2/3} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]{}^2+a^{2/3}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \sqrt[3]{k} \text{InverseFunction}\left[-\frac{\log \left(\text{$\#$1}^2 k^{2/3}+\text{$\#$1} \left(-\sqrt[3]{a}\right) \sqrt[3]{k}+a^{2/3}\right)-2 \log \left(\text{$\#$1} \sqrt[3]{k}+\sqrt[3]{a}\right)+2 \sqrt{3} \tan ^{-1}\left(\frac{1-\frac{2 \text{$\#$1} \sqrt[3]{k}}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 a^{2/3} \sqrt[3]{k}}\&\right][-t+c_1]}{\sqrt[3]{a}}}{\sqrt{3}}\right)}{6 \sqrt[3]{a} k^{2/3}}+c_2\right\}\right\}$$
 
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FAQ: Solution of a parametric differential equation

What is a parametric differential equation?

A parametric differential equation is an equation that involves one or more parameters, in addition to the independent and dependent variables. These parameters can be varied to obtain different solutions to the equation.

How is a parametric differential equation solved?

A parametric differential equation is solved by finding a set of values for the parameters that satisfies the equation. This can be done analytically or numerically using techniques such as separation of variables, substitution, or numerical integration.

What is the difference between a parametric differential equation and a non-parametric differential equation?

A parametric differential equation involves parameters that can be varied, while a non-parametric differential equation does not. This means that a parametric differential equation has an infinite number of solutions, while a non-parametric differential equation typically has a unique solution.

What are some real-world applications of parametric differential equations?

Parametric differential equations are used in many fields of science and engineering, such as physics, chemistry, biology, and economics. They can be used to model and understand complex systems, such as population growth, chemical reactions, and fluid dynamics.

Can a parametric differential equation have multiple solutions?

Yes, a parametric differential equation can have an infinite number of solutions, as the parameters can be varied to obtain different solutions. However, not all parametric differential equations have multiple solutions, and some may have a unique solution for a given set of parameters.

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