Question #20d0b

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Answer 1 · 2017-03-18T20:03:10

$y' + p(x)y=q(x)y^n$

Rewrite, $2x dy/dx = 10x^3 y^5 + y$ , in the above form:

$y' - 1/(2x)y = 5x^2y^5$

Divide both sides by y^5

$y'y^-5 - 1/(2x)y^-4 = 5x^2$

Let $u = y^-4$ , then $u' = 4y^-5y' or y'y^-5=(u')/4$ :

$(u')/4 - 1/(2x)u = 5x^2$

$u' - 2/(x)u = 20x^2" [1]"$

The integrating factor for equation [1] is $I = e^(int-2/xdx)$ which is a trivial integration:

$I = e^(-2ln(x))$

$I = e^ln(1/x^2)$

$I = 1/x^2$

Divide equation [1] by $x^2$

$(u')/x^2 - 2/(x^3)u = 20$

We know that the left side is the reverse of the product rule so integration is simple:

$(u(x))/x^2 = 20x+C$

Multiply both sides by $x^2$ :

$u(x) = 20x^3+Cx^2$

Reverse the substitution:

$y^-4=20x^3+Cx^2$

$y = (20x^3+Cx^2)^(-1/4)$