David Troyer

Answered

2021-12-26

For each of the following differential equations, determine the general or particular solution: $2x{y}^{\prime}+y={y}^{2}\mathrm{log}x$

Answer & Explanation

Mary Herrera

Expert

2021-12-27Added 37 answers

Step 1

Given:

Differential equation

$x{y}^{\prime}+y={y}^{2}\mathrm{log}x$

To find :

General solution of the given Differential equation?

Step 2

Solution :

We have given

$x{y}^{\prime}+y={y}^{2}\mathrm{log}x$

The given DE equation is a Bernoulli equation of the form

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right){y}^{n}$

And we can solve this kind of equation by substituting

$v={y}^{1-n}$

So we have the equation$x{y}^{\prime}+y={y}^{2}\mathrm{log}x$

and here the value of n is 22

So we will substitute

$v={y}^{1-2}$

$v=\frac{1}{y}$

Now differentiate with respect to x

$\frac{dv}{dx}=\frac{-1}{{y}^{2}}\frac{dy}{dx}$

$\frac{dy}{dx}=-{y}^{2}\frac{dv}{dx}$

Now we can write the given equation as

$x\frac{dy}{dx}+y={y}^{2}\mathrm{log}x$

$\frac{dy}{dx}+\frac{y}{x}=\frac{{y}^{2}\mathrm{log}x}{x}$

Now put the value of$\frac{dy}{dx}=-{y}^{2}\frac{dv}{dx}$

$-{y}^{2}\frac{dv}{dx}+\frac{y}{x}=\frac{{y}^{2}\mathrm{log}x}{x}$

divide both the sides by$-{y}^{2}$

$\frac{dv}{dx}-\frac{1}{xy}=\frac{-\mathrm{log}x}{x}$

Now put$v=\frac{1}{y}$

$\frac{dv}{dx}-\frac{v}{x}=\frac{-\mathrm{log}x}{x}$

Step 3

We got the equation

$\frac{dv}{dx}-\frac{v}{x}=\frac{-\mathrm{log}x}{x}$

Now we will find out the integrating factor for the above equation

$P\left(x\right)=\frac{-1}{x}\text{}div>}>$

Given:

Differential equation

To find :

General solution of the given Differential equation?

Step 2

Solution :

We have given

The given DE equation is a Bernoulli equation of the form

And we can solve this kind of equation by substituting

So we have the equation

and here the value of n is 22

So we will substitute

Now differentiate with respect to x

Now we can write the given equation as

Now put the value of

divide both the sides by

Now put

Step 3

We got the equation

Now we will find out the integrating factor for the above equation

0

esfloravaou

Expert

2021-12-28Added 43 answers

Rewrite Bernoulli Equation in the appropriate form:

$x{y}^{\prime}+y={y}^{2}\mathrm{ln}x$

$\frac{dy}{dx}+\frac{1}{x}y=\frac{\mathrm{ln}x}{x}{y}^{2}$

Use a substitution:

$v={y}^{1-2}={y}^{-1}$

$\frac{dv}{dx}=-{y}^{-2}\frac{dy}{dx}$

$=-{y}^{-2}(-\frac{1}{x}y+\frac{\mathrm{ln}x}{x}{y}^{2})$

$=\frac{1}{x}{y}^{-1}-\frac{\mathrm{ln}x}{x}$

$=\frac{1}{x}v-\frac{\mathrm{ln}x}{x}$

$\frac{dv}{dx}-\frac{1}{x}v=-\frac{\mathrm{ln}x}{x}$

Find integrating factor:

$\mu ={e}^{\int -\frac{1}{x}dx}={e}^{-\mathrm{ln}x}=\frac{1}{x}$

$\frac{1}{x}\frac{dv}{dx}-\frac{1}{{x}^{2}}v=-\frac{\mathrm{ln}x}{{x}^{2}}$

Solve for v:

$\frac{d}{dx}\left(\frac{v}{x}\right)=-\frac{\mathrm{ln}x}{{x}^{2}}$

$d\left(\frac{v}{x}\right)=-\frac{\mathrm{ln}x}{{x}^{2}}dx$

Integrate by parts:

$\frac{v}{x}=\int -\frac{\mathrm{ln}x}{{x}^{2}}dx$

$=\frac{1}{x}\mathrm{ln}x-\int \frac{1}{{x}^{2}}dx$

$=\frac{1}{x}\mathrm{ln}x+\frac{1}{x}+C$

$v=Cx+\mathrm{ln}x+1$

Substitute back and solve for y:

${y}^{-1}=Cx+\mathrm{ln}x+1$

Use a substitution:

Find integrating factor:

Solve for v:

Integrate by parts:

Substitute back and solve for y:

karton

Expert

2022-01-09Added 439 answers

You differentiate implicitly,

Rearranging

Factorising

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