Question

Particular integral of d²y/dx² +3dy/xdx +2/x² = 0 is

Answer 1

Let:

$y'=v(x)$

then:

$y''=v'$

Let:

$\mu(x)=e^{\int3/xdx}=x^3$

Multiply all terms of the given equation by $\mu(x)$ :

$x^3v'+3x^2v=-2x$

Substitute $3x^2=\frac{d}{dx}(x^3)$ :

$x^3\frac{dv(x)}{dx}+\frac{d}{dx}(x^3)v(x)=-2x$

Then:

$\frac{d}{dx}(x^3v)=-2x$

$\int\frac{d}{dx}(x^3v)dx=-\int2xdx$

$x^3v=-x^2+c_1$

$v=\frac{dy}{dx}=\frac{-x^2+c_1}{x^3}$

Answer:

$y(x)=\int\frac{-x^2+c_1}{x^3}dx=-lnx-\frac{c_1}{2x^2}+c_2$