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):

$x^5w'+5x^4w = \dfrac1x$

Bear in mind that in order to do this, we require $x\neq0$. 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,

$\left(x^5w\right)' = \dfrac1x$

Integrate both sides with respect to $x$ :

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

Solve for $w(x)$ :

$w = \dfrac{\ln(x)+C_1}{x^5}$

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

$v = \displaystyle \int \frac{\ln(x)+C_1}{x^5} \,\mathrm dx$

Integrate by parts:

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

Solve for $y_2(x)$ :

$\displaystyle \frac{y_2}{x^2} = -\frac{4\ln(x)+C_1}{16x^4}+C_2 \\\\ y_2 = -\frac{4\ln(x)+C_1}{16x^2} + C_2x^2$

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

$\displaystyle y_2 = -\frac{4\ln(x)+C_1}{16x^2}$

Still, the general solution would be

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