How to you find the general solution of $(2+x)y\prime =3y$?

excefebraxp
2022-09-12
Answered

How to you find the general solution of $(2+x)y\prime =3y$?

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Peyton Cox

Answered 2022-09-13
Author has **18** answers

We have;

$(2+x)\frac{dy}{dx}=3y$

This is a First Order separable Differential equation, and we can rearrange as follows:

$\frac{1}{y}\frac{dy}{dx}=\frac{3}{x+2}$

And we can "separate the variables" to get:

$\int \frac{1}{y}dy=\int \frac{3}{x+2}dx$

This is straightforward to integrate:

$\mathrm{ln}y=3\mathrm{ln}(x+2)+\mathrm{ln}A$

$={\mathrm{ln}(x+2)}^{3}+\mathrm{ln}A$

$=\mathrm{ln}A{(x+2)}^{3}$

$\therefore y=A{(x+2)}^{3}$

$(2+x)\frac{dy}{dx}=3y$

This is a First Order separable Differential equation, and we can rearrange as follows:

$\frac{1}{y}\frac{dy}{dx}=\frac{3}{x+2}$

And we can "separate the variables" to get:

$\int \frac{1}{y}dy=\int \frac{3}{x+2}dx$

This is straightforward to integrate:

$\mathrm{ln}y=3\mathrm{ln}(x+2)+\mathrm{ln}A$

$={\mathrm{ln}(x+2)}^{3}+\mathrm{ln}A$

$=\mathrm{ln}A{(x+2)}^{3}$

$\therefore y=A{(x+2)}^{3}$

asked 2022-05-15

I was wondering if I could get some advice on how to tackle this question:

Consider the differential equation

${x}^{2}\frac{dy}{dx}+2xy-{y}^{3}=0\phantom{\rule{1em}{0ex}}(3)$

Make the substitution $u={y}^{-2}$ and show that the differential equation reduces to

$-\frac{1}{2}{x}^{2}\frac{du}{dx}+2xu-1=0\phantom{\rule{1em}{0ex}}(4)$

Solve equation (4) for u(x) and hence write down the solution for equation (3).

I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:

$\begin{array}{rl}u& ={y}^{-2}\\ & =\frac{1}{{y}^{2}}\\ \therefore {y}^{2}& =\frac{1}{u}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y& =\pm \sqrt{\frac{1}{u}}\end{array}$

I'm not sure where to continue on from here though.

Consider the differential equation

${x}^{2}\frac{dy}{dx}+2xy-{y}^{3}=0\phantom{\rule{1em}{0ex}}(3)$

Make the substitution $u={y}^{-2}$ and show that the differential equation reduces to

$-\frac{1}{2}{x}^{2}\frac{du}{dx}+2xu-1=0\phantom{\rule{1em}{0ex}}(4)$

Solve equation (4) for u(x) and hence write down the solution for equation (3).

I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:

$\begin{array}{rl}u& ={y}^{-2}\\ & =\frac{1}{{y}^{2}}\\ \therefore {y}^{2}& =\frac{1}{u}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y& =\pm \sqrt{\frac{1}{u}}\end{array}$

I'm not sure where to continue on from here though.

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asked 2022-07-13

What is the standard method for finding solutions of differential equations such as this one? (if there is any)

$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

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What is a solution to the differential equation $x\frac{dy}{dx}=\frac{1}{y}$?

asked 2022-06-21

I have a differential equation of the form

$dy/dx+p(x)y=q(x)$

under the condition that $q(x)=300$ if $y<3312$ and $q(x)=0$ if $y\ge 3312$.

I could not understand how to solve this differential equation with such heavy side function ?

Any hints?

$dy/dx+p(x)y=q(x)$

under the condition that $q(x)=300$ if $y<3312$ and $q(x)=0$ if $y\ge 3312$.

I could not understand how to solve this differential equation with such heavy side function ?

Any hints?

asked 2022-09-11

How do you find the general solution of the differential equation $\frac{dy}{dx}=2{x}^{-3}$?