Question

What is the general solution of the differential equation `dy/dx + y = xy^3`?

Answer 1

See below.

Explanation:

Making the change of variable `y = 1/z` we have the new version

`dy/dx + y - xy^3=0 -> (x - z^2 + z z')/z^3=0` or

`x - z^2 + z z' = 0`

Now calling `xi = z^2` we have

`x-xi+1/2 xi'=0`

Solving for `xi` we obtain easily

`xi =1/2+x+C e^(2x) = z^2` then

`z = pm sqrt(1/2+x+C e^(2x) ) = 1/y` then finally

`y = pm 1/sqrt(1/2+x+C e^(2x) )`

Answer 2

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

Explanation:

Given: `dy/dx + y = xy^3`

Multiply both sides by `y^-3`:

`dy/dxy^-3 +y^-2 = x" [1]"`

Let `u = y^-2`, then `(du)/dx = -2y^-3dy/dx`

Writing the differential in a form that is suitable for substitution into equation [1]:

`-1/2(du)/dx = dy/dxy^-3`:

Perform the substitutions:

`-1/2(du)/dx + u = x`

Multiply the equation by -2:

`(du)/dx - 2u = -2x" [2]"`

This is the well known form:

`(dy)/dx +P(x)y = Q(x)`

Where `P(x) = -2`

The integrating factor is:

`mu(x) = e^(int-2dx)`

`mu(x) = e^(-2x)`

Multiply both sides of equation 2 by `mu(x)`:

`(du)/dxe^(-2x) - 2e^(-2x)u = -2xe^(-2x)`

We know that the left side integrates to `mu(x)u`, therefore, we need only integrate the right side:

`e^(-2x)u = -2intxe^(-2x)dx`

`e^(-2x)u = -2(xe^(-2x)/-2+1/2inte^(-2x)dx)`

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

`u = x+ 1/2+ ce^(2x)`

Reverse the substitution:

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

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

Answer 3

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

Alternatively:

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

Explanation:

We have:

`dy/dx + y = xy^3`

This is a Bernoulli equitation which has a standard method to solve. Let:

`u = y^(-2) => (du)/dy = -2y^(-3)` and `dy/(du) = -y^3/2`

By the chain rule we have;

`dy/dx = dy/(du) * (du)/dx`

Substituting into the last DE we get;

`dy/(du) (du)/dx + y = xy^3`
`:. -y^3/2 (du)/dx + y = xy^3`
`:. - (du)/dx + 2/y^2 = 2x`
`:. - (du)/dx + 2u = 2x`
`:. (du)/dx - 2u = -2x`

So the substitution has reduced the DE into a first order linear differential equation of the form:

`(d zeta)/dx + P(x) zeta = Q(x)`

We solve this using an Integrating Factor

`I = exp( \ int \ P(x) \ dx )`
`\ \ = exp( int \ (-2) \ dx )`
`\ \ = exp( -2x )`
`\ \ = e^(-2x)`

And if we multiply the last by this Integrating Factor, `I`, we will have a perfect product differential;

`e^(-2x)(du)/dx - 2ue^(-2x) = -2xe^(-2x)`
`d/dx(e^(-2x)u) = -2xe^(-2x)`

Which is now a trivial separable DE, so we can "separate the variables" to get:

`e^(-2x)u = int \ -2xe^(-2x) \ dx`

And integrating by parts (skipped step) gives us:

`e^(-2x)u = 1/2(2x+1)e^(-2x) + c`

Restoring the substitution we get:

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

`:. y^(-2) = 1/2(2x+1) + ce^(-2x)`
`:. 1/y^2 = 1/2(2x+1+Ce^(-2x))`
`:. y^2 = 2/(2x+1+Ce^(-2x))`