What is the general solution of the differential equation $\frac{dy}{dx}-2xy=x$?

Lindsey Ibarra
2022-09-27
Answered

What is the general solution of the differential equation $\frac{dy}{dx}-2xy=x$?

You can still ask an expert for help

brodireo1

Answered 2022-09-28
Author has **12** answers

$\frac{dy}{dx}-2xy=x$

Which we can write as:

$\frac{dy}{dx}=2xy+x$

$\therefore \frac{dy}{dx}=(2y+1)x$

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

Which is a first order separable differential equation, so we can "separate the variables" to get:

$\int \frac{1}{2y+1}dy=\int xdx$

Integrating we get, the General Solution:

$\frac{1}{2}\mathrm{ln}|2y+1|=\frac{1}{2}{x}^{2}+C$

Applying the initial condition y(1)=1 we find:

$\frac{1}{2}\mathrm{ln}3=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}\mathrm{ln}3-\frac{1}{2}$

So we can write an implicit particular solution as:

$\frac{1}{2}\mathrm{ln}|2y+1|=\frac{1}{2}{x}^{2}+\frac{1}{2}\mathrm{ln}3-\frac{1}{2}$

We typically require an explicit solution, so we can rearrange as follows:

$\mathrm{ln}|2y+1|={x}^{2}+\mathrm{ln}3-1$

$\therefore |2y+1|={e}^{{x}^{2}+\mathrm{ln}3-1}$

Noting that the exponential function is positive over its entire domain, (as $e}^{x}>0\forall x\in \mathbb{R$):

$2y+1={e}^{{x}^{2}-1}{e}^{\mathrm{ln}3}$

$\therefore 2y=3{e}^{{x}^{2}-1}-1$

$\therefore y=\frac{3}{2}{e}^{{x}^{2}-1}-\frac{1}{2}$

Which we can write as:

$\frac{dy}{dx}=2xy+x$

$\therefore \frac{dy}{dx}=(2y+1)x$

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

Which is a first order separable differential equation, so we can "separate the variables" to get:

$\int \frac{1}{2y+1}dy=\int xdx$

Integrating we get, the General Solution:

$\frac{1}{2}\mathrm{ln}|2y+1|=\frac{1}{2}{x}^{2}+C$

Applying the initial condition y(1)=1 we find:

$\frac{1}{2}\mathrm{ln}3=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}\mathrm{ln}3-\frac{1}{2}$

So we can write an implicit particular solution as:

$\frac{1}{2}\mathrm{ln}|2y+1|=\frac{1}{2}{x}^{2}+\frac{1}{2}\mathrm{ln}3-\frac{1}{2}$

We typically require an explicit solution, so we can rearrange as follows:

$\mathrm{ln}|2y+1|={x}^{2}+\mathrm{ln}3-1$

$\therefore |2y+1|={e}^{{x}^{2}+\mathrm{ln}3-1}$

Noting that the exponential function is positive over its entire domain, (as $e}^{x}>0\forall x\in \mathbb{R$):

$2y+1={e}^{{x}^{2}-1}{e}^{\mathrm{ln}3}$

$\therefore 2y=3{e}^{{x}^{2}-1}-1$

$\therefore y=\frac{3}{2}{e}^{{x}^{2}-1}-\frac{1}{2}$

asked 2022-06-18

I have the following differential equation problem but I couldn't proceed any further -

$\frac{dy}{dx}=\frac{\frac{-a\text{}y}{x}}{1-\text{}b\text{}{\left(\frac{1}{x}\right)}^{n}{\left(\frac{y}{1-y}\right)}^{m}}$

where, $x\in [0,1]\text{}\text{and}\text{}y\in [0,1]$

But I can't solve it down.

I have tried $y=u{y}_{1}\text{}\text{}\text{}\text{where},\text{}{y}_{1}={x}^{\frac{n}{m}}\text{}$, but it didn't help.

Wolfram gives the solution as -

$y(x)={c}_{1}\mathrm{exp}(\int \frac{a}{x-b\frac{{x}^{m-n+1}}{(-x+1{)}^{m}}}\phantom{\rule{thinmathspace}{0ex}}dx)$

How to simplify the integral?

I just wanted a hint that whether it can be solved? If yes, please just tell me what am I doing wrong.

$\frac{dy}{dx}=\frac{\frac{-a\text{}y}{x}}{1-\text{}b\text{}{\left(\frac{1}{x}\right)}^{n}{\left(\frac{y}{1-y}\right)}^{m}}$

where, $x\in [0,1]\text{}\text{and}\text{}y\in [0,1]$

But I can't solve it down.

I have tried $y=u{y}_{1}\text{}\text{}\text{}\text{where},\text{}{y}_{1}={x}^{\frac{n}{m}}\text{}$, but it didn't help.

Wolfram gives the solution as -

$y(x)={c}_{1}\mathrm{exp}(\int \frac{a}{x-b\frac{{x}^{m-n+1}}{(-x+1{)}^{m}}}\phantom{\rule{thinmathspace}{0ex}}dx)$

How to simplify the integral?

I just wanted a hint that whether it can be solved? If yes, please just tell me what am I doing wrong.

asked 2022-09-29

What is a particular solution to the differential equation $\frac{dy}{dx}={e}^{x-y}$ with y(0)=2?

asked 2022-09-29

How do you solve for xy'−y=3xy given y(1)=0?

asked 2022-04-10

For every differentiable function $f:\mathbb{R}\to \mathbb{R}$, there is a function $g:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ such that $g(f(x),{f}^{\prime}(x))=0$ for every $x$ and for every differentiable function $h:\mathbb{R}\to \mathbb{R}$ holds that

being true that for every $x\in \mathbb{R}$, $g(h(x),{h}^{\prime}(x))=0$ and $h(0)=f(0)$ implies that $h(x)=f(x)$ for every $x\in \mathbb{R}$.

i.e every differentiable function $f$ is a solution to some first order differential equation that has translation symmetry.

being true that for every $x\in \mathbb{R}$, $g(h(x),{h}^{\prime}(x))=0$ and $h(0)=f(0)$ implies that $h(x)=f(x)$ for every $x\in \mathbb{R}$.

i.e every differentiable function $f$ is a solution to some first order differential equation that has translation symmetry.

asked 2022-06-01

How can I start to solve this differential equation?

${y}^{\prime}=\frac{y}{2y\mathrm{ln}(y)+y-x}$

${y}^{\prime}=\frac{y}{2y\mathrm{ln}(y)+y-x}$

asked 2020-11-26

Solve differential equation

asked 2021-03-07

Solve differential equation$x{u}^{\prime}(x)={u}^{2}-4$