Question

Can someone help me solving this differential equation?

Find the general solution of [tex]x^4y''+x^3y'-4x^2y=1 given tha [tex]t y_1=x^2 is a solution of the associated homogeneous equation.

Answer 1

Use reduction of order. Given a solution
`y_1(x) = x^2`, look for a second solution of the form
`y_2(x) = y_1(x)v(x)`.

Compute the first two derivatives of
`y_2(x)`:

`y_2 = x^2v \\\\ {y_2}' = x^2v' + 2xv \\\\ {y_2}'' = x^2v''+4xv' + 2v`

Substitute them into the ODE:

`x^4 (x^2v'' + 4xv' + 2v) + x^3 (x^2v' + 2xv) - 4x^2 (x^2v) = 1 \\\\ x^6v'' + 5x^5v' = 1`

Now substitute
`w(x) = v'(x)` and you end up with a linear ODE:

`x^6w'+5x^5w=1`

Multiply through both sides by
`\frac1x` (if you're familiar with the integrating factor method, this is it):

`x^5w'+5x^4w = \frac1x`

Bear in mind that in order to do this, we require
`x\\eq0`. Just to avoid having to deal with absolute values later, let's further assume
`x>0`.

Notice that the left side is the derivative of a product,

`\left(x^5w\right)' = \frac1x`

Integrate both sides with respect to
`x` :

`x^5w = \displaystyle \int\frac{\mathrm dx}x \\\\ x^5w = \ln(x) + C_1`

Solve for
`w(x)` :

`w = (\ln(x)+C_1)/(x^5)`

Solve for
`v(x)` by integrating both sides:

`v = \displaystyle \int (\ln(x)+C_1)/(x^5) \,\mathrm dx`

Integrate by parts:

`\displaystyle f = \ln(x) + C_1 \implies \mathrm df = \frac{\mathrm dx}x \\\\ \mathrm dg = (\mathrm dx)/(x^5) \implies g = -\frac1{4x^4} \\\\ \implies v = -(\ln(x)+C_1)/(4x^4) + \frac14 \int (\mathrm dx)/(x^5) \\\\ v = -(\ln(x)+C_1)/(4x^4) - \frac1{16x^4} + C_2 \\\\ v = -(4\ln(x)+C_1)/(16x^4)+C_2`

Solve for
`y_2(x)` :

`\displaystyle (y_2)/(x^2) = -(4\ln(x)+C_1)/(16x^4)+C_2 \\\\ y_2 = -(4\ln(x)+C_1)/(16x^2) + C_2x^2`

But since
`y_1(x)=x^2` is already accounted for, the second solution is just

`\displaystyle y_2 = -(4\ln(x)+C_1)/(16x^2)`

Still, the general solution would be

`\displaystyle \boxed{y(x) = -(4\ln(x)+C_1)/(16x^2) + C_2x^2}`