Question

Solve the first order linear inhomogeneous differential equation using the Bernoulli method:

y^,+(3y/x)=(1/x³)

Answer 1

$$y'+\dfrac{3}{x}y=\dfrac{1}{x^3}$$

Let $v(x)=\dfrac{1}{x^2}$

$$v'(x)=-\dfrac{2}{x^3}$$

$$-\dfrac{2}{x^3}+\dfrac{3}{x^3}=\dfrac{1}{x^3}$$

The function $v(x)=\dfrac{1}{x^2}$ is the solution of the given differential equation.

Let $y=u(x)v(x)=u(x)(\dfrac{1}{x^2}).$ Then

$$y'=\dfrac{u'}{x^2}-\dfrac{2u}{x^3}$$

Substitute

$$\dfrac{u'}{x^2}-\dfrac{2u}{x^3}+\dfrac{3u}{x^3}=\dfrac{1}{x^3}$$

$$xu'+u=1$$

$$\dfrac{du}{u-1}=-\dfrac{dx}{x}$$

Integrate

$$\int \dfrac{du}{u-1}=-\int\dfrac{dx}{x}$$

$$u-1=\dfrac{C}{x}$$

$$y=(\dfrac{C}{x}+1)(\dfrac{1}{x^2})$$

$$y(x)=\dfrac{1}{x^2}+\dfrac{C}{x^3}$$