Question

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

Hint: try a substitution `z = 1/y^2`

Answer 1

`y^2 = 2/(2x + 1 + Ae^(2x) )`

Explanation:

We have:

`dy/dx + y = xy^3` ..... [A]

As suggested we perform a substitution:

`z = 1/y^2 iff y^2 = 1/z`

The differentiating wrt `x` we have:

`(dz)/(dy) = -2/y^3 => (dy)/(dz) =-y^3/2`

And from the chain rule, we have:

`dy/dx = dy/dz * dz/dx = -y^3/2 dz/dx`

Substituting into the initial Differential Equation [A], we have:

`-y^3/2 dz/dx + y = xy^3`

Dividing by `y^3` we have:

`-1/2 dz/dx + 1/y^2 = x`

`:. -1/2 dz/dx + 1/(1/z) = x`

`:. dz/dx -2z = -2x` ..... [B}

This substitution has reduced the equation [A] to a Ordinary Differential Equation of the form, which can be solved using an integrating Factor;

`dz/dx + P(x)z=Q(x)`

Then the integrating factor is given by;

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

And if we multiply the DE [B] by this Integrating Factor, `I`, we will have a perfect product differential;

`dz/dx e^(-2x) -2e^(-2x) z= -2xe^(-2x)`
`:. d/dx( e^(-2x)z) = -2xe^(-2x)`

We can now "separate the variables", to get:

`e^(-2x)z = -2 \ int \ xe^(-2x) \ dx` ..... [C]

To integrate this integral we will require an application of Integration By Parts:

Let `{ (u,=x, => (du)/dx,=1), ((dv)/dx,=e^(-2x), => v,=-1/2e^(-2x) ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us

`int \ (x)(e^(-2x)) \ dx = (x)(-1/2e^(-2x)) - int \ (-1/2e^(-2x))(1) \ dx`
`:. int \ xe^(-2x) \ dx = -1/2 xe^(-2x) + 1/2 int \ e^(-2x) \ dx`
`:. int \ xe^(-2x) \ dx = -1/2 xe^(-2x) -1/4 e^(-2x)`

Using this result, we can now integrate [C] to get:

`e^(-2x)z = -2 {-1/2 xe^(-2x) -1/4 e^(-2x) } +c`

`:. ze^(-2x) = xe^(-2x) +1/2 e^(-2x) +c`

`:. z = x +1/2 +c e^(2x)`

Restoring the substitution we get:

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

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

Which we can write as:

`y^2 = 2/(2x + 1 + Ae^(2x) )`

Which is the general solution, or:

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

The initial conditions are invalid as `y` is given as a function, rather than a value!