Question #351040

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

Expert's answer

"y=uv\\\\\ny'=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}"
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