Solve the Differential Equation  x^2y'' -xy'-8y=0 ?

Jul 3, 2017

$y \left(x\right) = {c}_{1} / {x}^{2} + {c}_{2} {x}^{4}$

Explanation:

${x}^{2} y ' ' - x y ' - 8 y = 0$

This is a Cauchy-Euler differential equation of second order, so the resolution method is to substitute as possible solution:

$y \left(x\right) = {x}^{n}$

and its derivatives:

$y ' \left(x\right) = n {x}^{n - 1}$

$y ' ' \left(x\right) = n \left(n - 1\right) {x}^{n - 2}$

Substituting in the original equation we have:

x^2((n(n-1)x^(n-2))-nx x^(n-1)-8x^n =0

(n(n-1)x^n-nx^n-8x^n =0

${x}^{n} \left({n}^{2} - n - n - 8\right) = 0$

${x}^{n} \left({n}^{2} - 2 n - 8\right) = 0$

So for $x \ne 0$ we must have:

${n}^{2} - 2 n - 8 = 0$

$n = \left(1 \pm \sqrt{1 + 8}\right)$

${n}_{1} = - 2$
${n}_{2} = 4$

The general solution is then:

$y \left(x\right) = {c}_{1} / {x}^{2} + {c}_{2} {x}^{4}$

Jul 3, 2017

$y = \frac{A}{x} ^ 2 + B {x}^{4}$

Explanation:

We have:

${x}^{2} y ' ' - x y ' - 8 y = 0$ ..... [A]

This is a Euler-Cauchy Equation which is typically solved via a change of variable. Consider the substitution:

$x = {e}^{t} \implies x {e}^{- t} = 1$

Then we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- t} \frac{\mathrm{dy}}{\mathrm{dt}}$, and, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t}$

Substituting into the initial DE [A] we get:

${x}^{2} \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t} - x {e}^{- t} \frac{\mathrm{dy}}{\mathrm{dt}} - 8 y = 0$

$\therefore \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) - \frac{\mathrm{dy}}{\mathrm{dt}} - 8 y = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - 2 \frac{\mathrm{dy}}{\mathrm{dt}} - 8 = 0$ ..... [B]

This is now a second order linear homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

${m}^{2} - 2 m - 8 = 0$

We can solve this quadratic equation, and we get two real and distint roots:

$m = - 2 , 4$

Thus the Homogeneous equation [B] has the solution:

$y = A {e}^{- 2 t} + B {e}^{4 t}$

Now we initially used a change of variable:

$x = {e}^{t} \implies t = \ln x$

So restoring this change of variable we get:

$y = A {e}^{- 2 \ln x} + B {e}^{4 \ln x}$

$\therefore y = A {e}^{\ln {x}^{- 2}} + B {e}^{\ln {x}^{4}}$
$\therefore y = A {x}^{- 2} + B {x}^{4}$
$\therefore y = \frac{A}{x} ^ 2 + B {x}^{4}$

Which is the General Solution.