Question

Does anybody know how to solve this?.....yy''=y^2y'+y'^2

Answer 1

`y =( C e^(C(x + D)))/(1 - e^(C(x + D)))`

and

`y = " const"`

Explanation:

`yy''=y^2y'+y'^2`

`y = " const"` is obviously a trivial solution as every term has a derivative in it.

Taking it further, a common physics trick:

• `y'' = (d^2 y)/(dx^2) = (d y') /(dx) = (dy')/(dy) * (dy)/(dx) = (dy')/(dy) y'`

Substituting:

`y(dy')/(dy) y'=y^2y'+y'^2`

`(dy')/(dy) - (y')/y =y`

Integrating factor `bb I (y)`:

`bb I ( y) = exp (int \- 1/y \ dy) = exp (int \- ln y + ln C) = C/y`

`bb I ((dy')/(dy) - (y')/y =y )`

`implies 1/y (dy')/(dy) - (y')/y^2 = 1`

`( 1/y y')^' = 1`

`1/y y' = y + C`

`y' = y(y + C)`

This is separable:

`dy/(y^2 + C y) = dx`

`1/C ln(y/(y + C)) = x + D`

`y/(y + C) = e^(C(x + D))`

`y (1 - e^(C(x + D))) = C e^(C(x + D))`

`y =( C e^(C(x + D)))/(1 - e^(C(x + D)))`