I am wondering how to find a general analytical solution to the following ODE:

The solution method might be relatively simple; but right now I don't know how to approach this problem.

adiadas8o7
2022-04-30
Answered

Second-order ODE involving two functions

I am wondering how to find a general analytical solution to the following ODE:

$\frac{dy}{dt}\frac{{d}^{2}x}{{dt}^{2}}=\frac{dx}{dt}\frac{{d}^{2}y}{{dt}^{2}}$

The solution method might be relatively simple; but right now I don't know how to approach this problem.

I am wondering how to find a general analytical solution to the following ODE:

The solution method might be relatively simple; but right now I don't know how to approach this problem.

You can still ask an expert for help

Felicity Carter

Answered 2022-05-01
Author has **16** answers

Step 1

I find it more convenient to rewrite the equation using Newton's notation. Instead of writing$\frac{dx}{dt}$ , it is more helpful to write x'. Thus, the equation is

${x}^{\prime}y{}^{\u2033}=x{}^{\u2033}{y}^{\prime}.$

Now, suppose${x}^{\prime}=0$ . Then $x0$ is trivial, so every differentiable function $y\in {\mathbb{R}}^{\mathbb{R}}$ satisfies the equation. This holds analogously if ${y}^{\prime}=0$ . Otherwise, we can divide by x'y', thus

$\frac{y{}^{\u2033}}{{y}^{\prime}}=\frac{x{}^{\u2033}}{{x}^{\prime}}$ .

Step 2

There are four cases to consider from here:${x}^{\prime}<0\text{}\text{and}\text{}{y}^{\prime}0;{x}^{\prime}0\text{}\text{and}\text{}{y}^{\prime}0;{x}^{\prime}0\text{}\text{and}\text{}{y}^{\prime}0$ ; and ${x}^{\prime}>0\text{}\text{and}\text{}{y}^{\prime}0$ . These cases simplify the equation respectively

$\mathrm{ln}(-{x}^{\prime})+C=\mathrm{ln}(-{y}^{\prime})$

$\mathrm{ln}(-{x}^{\prime})+C=\mathrm{ln}\left({y}^{\prime}\right)$

$\mathrm{ln}\left({x}^{\prime}\right)+C=\mathrm{ln}(-{y}^{\prime})$

$\mathrm{ln}\left({x}^{\prime}\right)+C=\mathrm{ln}\left({y}^{\prime}\right)$

which are equivalent to

$e}^{C}{x}^{\prime}={y}^{\prime$

$-{e}^{C}{x}^{\prime}={y}^{\prime}$

$-{e}^{C}{x}^{\prime}={y}^{\prime}$

$e}^{C}{x}^{\prime}={y}^{\prime$ .

These cases simply reduce to$y}^{\prime}=A{x}^{\prime$

where$A\ne 0$ . Therefore, we have that $y\left(t\right)=Ax\left(t\right)+B$ , where $A\ne 0$ . Remember that this is in the case when neither x nor y is a constant function.

I find it more convenient to rewrite the equation using Newton's notation. Instead of writing

Now, suppose

Step 2

There are four cases to consider from here:

which are equivalent to

These cases simply reduce to

where

asked 2021-01-02

Find

asked 2022-05-01

Prove instability using Lyapunov function

${x}^{\prime}={x}^{3}+xy$

$y}^{\prime}=-y+{y}^{2}+xy-{x}^{3$

asked 2022-03-29

I have to solve the following Cauchy's problem:

$\{\begin{array}{rl}& {x}^{2}{x}^{\prime}={\mathrm{sin}}^{2}({x}^{3}-3t)\\ & x(0)=1\end{array}$

asked 2022-04-02

Doubt regarding a differential equation.

Solve:$x,dx+y,dy=\frac{{a}^{2}(x,dy-y,dx)}{{x}^{2}+{y}^{2}}$

The author further proceeds to rearrange above in the form$M,dx+N,dy=0$ where

$M=x+\frac{{a}^{2}y}{{x}^{2}+{y}^{2}};N=y-\frac{{a}^{2}x}{{x}^{2}+{y}^{2}}$

Which further implies

$M}_{y}={N}_{x}=\frac{{a}^{2}({x}^{2}-{y}^{2})}{{({x}^{2}+{y}^{2})}^{2}$

This is the required condition for the given equation to be exact and the solution is obtained using standard formula.

But, what I did is as follows:

$x,dx+y,dy=\frac{{a}^{2}(x,dy-y,dx)}{{x}^{2}+{y}^{2}}$

$\to ({x}^{2}+{y}^{2})x,dx+({x}^{2}+{y}^{2})y,dy={a}^{2}(x,dy-y,dx)$

Therefore$({x}^{3}+x{y}^{2}+{a}^{2}y),dx+({y}^{3}+y{x}^{2}-{a}^{2}x),dy=0$

Comparing it with the equation$M,dx+N,dy=0$ we get,

$M={x}^{3}+x{y}^{2}+{a}^{2}y;N={y}^{3}+y{x}^{2}-{a}^{2}x$

But$M}_{y}=2xy+{a}^{2};{N}_{x}=2xy-{a}^{2$

Solve:

The author further proceeds to rearrange above in the form

Which further implies

This is the required condition for the given equation to be exact and the solution is obtained using standard formula.

But, what I did is as follows:

Therefore

Comparing it with the equation

But

asked 2022-04-09

Use the annihilator method and find the general solution of $y{}^{\u2033}-3{y}^{\prime}+2y=4{\mathrm{sin}}^{3}\left(3x\right)$

asked 2022-04-10

Trying to understand eigenvalues with respect to differential equations.

I am trying to understand how to find eigenvalues from a matrix consisting of exponential terms, considering a differential equation. The examples I've seen online are ODEs. Without using a vector with exponential terms, here is what I have learned.

$\frac{d}{dt}\overrightarrow{x}\left(t\right)=\lambda \overrightarrow{x}\left(t\right)$

$\frac{d}{dt}\left[\begin{array}{c}{x}_{1}\left(t\right)\\ {x}_{2}\left(t\right)\\ {x}_{3}\left(t\right)\end{array}\right]=\left[\begin{array}{ccc}{\lambda}_{1}& 0& 0\\ 0& {\lambda}_{2}& 0\\ 0& 0& {\lambda}_{3}\end{array}\right]\left[\begin{array}{c}{x}_{1}(t=0)\\ {x}_{2}(t=0)\\ {x}_{3}(t=0)\end{array}\right]$

Here is what I am trying to understand

Section of a paper with imag eigenvalue

In this paper, the following assumption is made.

$\overrightarrow{J}\left(t\right)=\left|\overrightarrow{J}\right|{e}^{-i\omega t}=\left[\begin{array}{c}\left|{J}_{x}\right|{e}^{-i\omega t}\\ \left|{J}_{y}\right|{e}^{-i\omega t}\\ \left|{J}_{z}\right|{e}^{-i\omega t}\end{array}\right]$

They are using partial derivatives (I believe this can be viewed as an ODE then?). Differentiating with respect to time I believe yields the following. Please correct me if I am wrong.

$\frac{\partial}{\partial t}\overrightarrow{J}\left(t\right)=\frac{\partial}{\partial t}\left[\begin{array}{c}\left|{J}_{x}\right|{e}^{-i\omega t}\\ \left|{J}_{y}\right|{e}^{-i\omega t}\\ \left|{J}_{z}\right|{e}^{-i\omega t}\end{array}\right]=\left[\begin{array}{ccc}-i\omega & 0& 0\\ 0& -i\omega & 0\\ 0& 0& -i\omega \end{array}\right]\left[\begin{array}{c}\left|{J}_{x}\right|{e}^{-i\omega t(t=0)}\\ \left|{J}_{y}\right|{e}^{-i\omega t(t=0)}\\ \left|{J}_{z}\right|{e}^{-i\omega t(t=0)}\end{array}\right]$

Or

$\frac{\partial}{\partial t}\left[\begin{array}{c}\left|{J}_{x}\right|{e}^{-i\omega t}\\ \left|{J}_{y}\right|{e}^{-i\omega t}\\ \left|{J}_{z}\right|{e}^{-i\omega t}\end{array}\right]=\left[\begin{array}{ccc}-i\omega & 0& 0\\ 0& -i\omega & 0\\ 0& 0& -i\omega \end{array}\right]\left[\begin{array}{c}\left|{J}_{x}\right|\\ \left|{J}_{y}\right|\\ \left|{J}_{z}\right|\end{array}\right]$

Are my assumptions correct? If so, is there a deeper analysis to why this is the case with exponential terms?

asked 2022-04-13

Find the number of unique solutions of the following nonlinear ODE

${y}^{\prime}=\frac{10}{3}x{y}^{\frac{25}{,}}{\textstyle \phantom{\rule{2em}{0ex}}}y\left(0\right)=-1$