Question

Question #20d0b

Answer 1

This is a Bernoulli Differential Equation of the form :

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

Explanation:

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)`