Question

What is a solution to the differential equation `dy/dx=2yx+yx^2`?

Answer 1

`=alpha e ^(x^2(1 + x/3) )`

Explanation:

`dy/dx=2yx+yx^2`

separate it, starting with....

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

`1/y dy/dx=(2x+x^2)`

you might recognise `1/y dy/dx` as `(ln y)'` and so the integration can be done pretty infomally, but if not....we can integrate both sides as follows.....starting again with the last equation:

`1/y dy/dx=(2x+x^2)`

`1/y dy=(2x+x^2) \ dx`

`int dy qquad 1/y = int dx qquad (2x+x^2)`

`ln y = x^2 + x^3/3 + C`

`= x^2(1 + x/3) + C`

`implies y = e^ (x^2(1 + x/3) + C)`

`= e^ (x^2(1 + x/3) )*e^(C)`

`=alpha e ^(x^2(1 + x/3) )`

where `alpha = const = e^C`

Answer 2

`y = C_0 e^{x^2 + 1/3 x^3}`

Explanation:

Grouping variables

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

integrating both sides

`log_e y = x^2+1/3x^3 + C`

Finally

`y = C_0 e^{x^2 + 1/3 x^3}`