 David Troyer

2021-12-26

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

Expert

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 esfloravaou

Expert

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:
$v={y}^{1-2}={y}^{-1}$
$\frac{dv}{dx}=-{y}^{-2}\frac{dy}{dx}$
$=-{y}^{-2}\left(-\frac{1}{x}y+\frac{\mathrm{ln}x}{x}{y}^{2}\right)$
$=\frac{1}{x}{y}^{-1}-\frac{\mathrm{ln}x}{x}$
$=\frac{1}{x}v-\frac{\mathrm{ln}x}{x}$
$\frac{dv}{dx}-\frac{1}{x}v=-\frac{\mathrm{ln}x}{x}$
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:
${y}^{-1}=Cx+\mathrm{ln}x+1$ karton

Expert

You differentiate implicitly, $y+xdy/dx+dy/dx=2ydy/dx×1/x$
Rearranging
$-xdy/dx-dy/dx+2ydy/xdx=y$
Factorising $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{y}{\frac{-x-1+2y}{x}}$

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