# Solve the Differential Equation  x^2y'' +11xy'+25y=0 ?

Jul 3, 2017

$y \left(x\right) = {c}_{1} {x}^{-} 5 + {c}_{2} {x}^{- 5} \ln x$

#### Explanation:

Substitute the variable:

$t = \ln x$

$y \left(x\right) = \phi \left(\ln x\right) = \phi \left(t\right)$

so that:

$y ' \left(x\right) = \frac{1}{x} \phi '$

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

substituting in the original equation:

${x}^{2} y ' ' + 11 x y ' + 25 y = 0$

$\phi ' ' - \phi ' + 11 \phi ' + 25 \phi = 0$

$\phi ' ' + 10 \phi ' + 25 \phi = 0$

This is a second order equation with constant coefficients, so we can solve the characteristic equation:

${\lambda}^{2} + 10 \lambda + 25 = 0$

$\left(\lambda + 5\right) = 0$

$\lambda = - 5$

so the general solution is:

$\phi \left(t\right) = {c}_{1} {e}^{- 5 t} + {c}_{2} t {e}^{- 5 t}$

and undoing the substitution:

$y \left(x\right) = \phi \left(\ln x\right) = {c}_{1} {e}^{- 5 \ln x} + {c}_{2} \ln x {e}^{- 5 \ln x}$

$y \left(x\right) = {c}_{1} {x}^{-} 5 + {c}_{2} {x}^{- 5} \ln x$

Jul 3, 2017

$y = \left(A \ln x + B\right) {x}^{- 5}$

#### Explanation:

We have:

${x}^{2} y ' ' + 11 x y ' + 25 y = 0$ ..... [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:

${x}^{2} \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t} + 11 x {e}^{- t} \frac{\mathrm{dy}}{\mathrm{dt}} + 25 y = 0$

$\therefore \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) + 11 \frac{\mathrm{dy}}{\mathrm{dt}} + 25 y = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} + 10 \frac{\mathrm{dy}}{\mathrm{dt}} + 25 y = 0$ ..... [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.

${m}^{2} + 10 m + 25 = 0$

We can solve this quadratic equation, and we get a real repeated root:

$m = - 5$

Thus the Homogeneous equation [B] has the solution:

$y = \left(A x + B\right) {e}^{- 5 t}$

Now we initially used a change of variable:

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

So restoring this change of variable we get:

$y = \left(A \ln x + B\right) {e}^{- 5 \ln x}$

$\therefore y = \left(A \ln x + B\right) {x}^{- 5}$

Which is the General Solution.