Question

How can I solve this differential equation? x dy/dx+y=x^2y^2

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

Answer 1

`y=1/(Cx-x^2)`

Explanation:

This equation is a Bernoulli equation as it can be rewritten to the form `("d"y)/("d"x)+f(x)y=g(x)y^n`.

From `x ("d"y)/("d"x)+y=x^2y^2`, one can divide both sides by `x` so that it fits the Bernoulli form.

`("d"y)/("d"x)+y/x=xy^2`

Divide both sides by `y^2`:
`y^-2 ("d"y)/("d"x)+1/(xy)=x`

Then, define a function `v=y^(1-2)=y^-1`. Differentiate both sides with respect to `x` to get `("d"v)/("d"x)=-("d"y)/("d"x)y^-2`, or `("d"y)/("d"x)=-y^2("d"v)/("d"x)`. We can now substitute these values into the original ODE:
`-y^2*("d"v)/("d"x)*y^-2+v/x=x`
`-("d"v)/("d"x)+v/x=x`
`("d"v)/("d"x)-v/x=-x`

This now becomes a simple linear first-order ODE. To solve this, we multiply both sides by the integrating factor `e^(-int("d"x)/x)=1/x`:

`1/x("d"v)/("d"x)-v/x^2=-1`
`"d"/("d"x)(v/x)=-1`
`v/x=C-x`
`v=Cx-x^2`

Recall that we previously defined `v=y^-1`, or the solution `y=v^-1`.

Thus, our final answer is `y=1/(Cx-x^2)`.