Question

How do you solve `dy/dx = xy^2`?

Answer 1

`y(x)=-2/(x^2+C)`

Explanation:

Let's separate our variables, IE, have each side of the equation only in terms of one variable. This entails

`dy/y^2=xdx`

Integrate each side:

`intdy/y^2=intxdx`

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

Note that we would technically have constants of integration on both sides, but we moved them all over to the right and absorbed them into `C.`

Now, let's get an explicit solution with `y` as a function of `x:`

`-1=1/2x^2y+Cy`

`y(1/2x^2+C)=-1`

`y=-1/(1/2x^2+C)`

Let's get the fraction out of the denominator. It just looks messy.

`y=-1/((x^2+2C)/2`

Well, `2C=C,` as `2` multiplied by a constant just yields another constant.

`y=-1/((x^2+C)/2)`

Thus,

`y(x)=-2/(x^2+C)`

Answer 2

`y = -2/(x^2+C)`

where `C` is an arbitrary constant.

Explanation:

`dy/dx = xy^2`

We separate variables:

`dy/y^2 = xdx`

Now we integrate both sides:

`\int1/y^2dy = \intxdx`

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

where `C` is an arbitrary constant of integration.

Now we solve for `y`.

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

`y = -1/(1/2x^2+C)`

`=>y = -2/(x^2+C)`

where `C` absorbed a factor of `2` since it is an arbitrary constant.