Answer to Question #351040 in Differential Equations for Taras

Question

2.1. Solve 2xy + 6x + (x^2 - 4)y'=0

Answer 1

$$(x^2 - 4)y'+2xy=-6x$$

$$y=uv\\ y'=u'v+uv'$$

$$(x^2-4)(u'v+uv')+2xuv=-6x$$

$$u((x^2-4)v'+2xv)+(x^2-4)vu'=-6x$$

Let

$$(x^2-4)v'+2xv=0$$

Then

$$\frac{dv}{v}=-\frac{2x}{x^2-4}dx$$

$$\ln v=-\ln (x^2-4)$$

$$v=\frac{1}{x^2-4}$$

We get an equation

$$(x^2-4)\frac{1}{x^2-4}u'=-6x$$

$$u'=-6x$$

$$u=-3x^2+C$$

Finally

$$y=uv=\frac{-3x^2+C}{x^2-4}$$