Question

How do you solve `y'=(2x-3)/(y^2+y+4)` given y(1)=2?

Answer 1

`2y^3+3y^2+24y = 6x^2-18x + 44`

Explanation:

We have `y'=(2x-3)/(y^2+y+4)`, or

`dy/dx=(2x-3)/(y^2+y+4)`

Which is a simple First Order separable DE, we can therefore separate the variables to get:
`(y^2+y+4)dy/dx=(2x-3)`
`:. int (y^2+y+4)dy = int (2x-3)dx`

We can integrate this to get:

`1/3y^3+1/2y^2+4y = x^2-3x + C`

Given `y(1)=2` we get:

`1/3 2^3+1/2 2^2+4*2 = 1^2-3 + C`
`:. 8/3+2+8 = 1-3 + C`
`:. C=44/3`

So the solution is:

`1/3y^3+1/2y^2+4y = x^2-3x + 44/3`
`:. 2y^3+3y^2+24y = 6x^2-18x + 44`