# What is the general solution of the differential equation?  x^2y'' +3xy'+17y=0

Jul 3, 2017

$y = \frac{{C}_{1} \cos \left(4 \log x\right) + {C}_{2} \sin \left(4 \log x\right)}{x}$

#### Explanation:

Assuming that the differential equation reads

${x}^{2} y ' ' + 3 x y ' + 17 y = 0$

proposing a solution with the structure

$y = c {x}^{\alpha}$ and substituting

${x}^{2} \alpha \left(\alpha - 1\right) c {x}^{\alpha - 2} + 3 \alpha c {x}^{\alpha - 1} + 17 c {x}^{\alpha} = \left(\alpha \left(\alpha - 1\right) + 3 \alpha + 17\right) c {x}^{\alpha} = 0$

Solving now

$\alpha \left(\alpha - 1\right) + 3 \alpha + 17 = 0$ we obtain

$\alpha = - 1 \pm 4 i$ so the solutions are

$y = {c}_{1} {x}^{- 1 - 4 i} + {c}_{2} {x}^{- 1 + 4 i}$

but

${x}^{4 i} = {e}^{i 4 \log x}$ and

${e}^{i 4 \log x} = \cos \left(4 \log x\right) + i \sin \left(4 \log x\right)$

so we can reduce the solutions to the form

$y = \frac{{C}_{1} \cos \left(4 \log x\right) + {C}_{2} \sin \left(4 \log x\right)}{x}$

Jul 3, 2017

$y = \frac{A \cos \left(4 \ln x\right)}{x} + \frac{B \sin \left(4 \ln x\right)}{x}$

#### Explanation:

We have:

${x}^{2} y ' ' + 3 x y ' + 17 y = 0$ ..... [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:

${x}^{2} \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t} + 3 x {e}^{- t} \frac{\mathrm{dy}}{\mathrm{dt}} + 17 y = 0$

$\therefore \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) + 3 \frac{\mathrm{dy}}{\mathrm{dt}} + 17 y = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} + 2 \frac{\mathrm{dy}}{\mathrm{dt}} + 17 y = 0$ ..... [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.

${m}^{2} + 2 m + 17 = 0$

We can solve this quadratic equation, and we get two complex conjugate roots:

$m = - 1 \pm 4 i$

Thus the Homogeneous equation [B] has the solution:

$y = {e}^{- t} \left(A \cos 4 t + B \sin 4 t\right)$

Now we initially used a change of variable:

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

So restoring this change of variable we get:

$y = \left({x}^{- 1}\right) \left(A \cos \left(4 \ln x\right) + B \sin \left(4 \ln x\right)\right)$

$\therefore y = \frac{A \cos \left(4 \ln x\right)}{x} + \frac{B \sin \left(4 \ln x\right)}{x}$

Which is the General Solution.