# Solve the Differential Equation  x^2y'' -3x+5y=0 ?

Jun 15, 2017

$y = \sqrt{x} \left(A \cos \left(\frac{1}{2} \sqrt{19} \ln x\right) + B \cos \left(\frac{1}{2} \sqrt{19} \ln x\right)\right) + \frac{3}{5} x$

#### Explanation:

We have:

${x}^{2} y ' ' - 3 x + 5 y = 0$

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 we get:

${x}^{2} \left(\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}}\right) {e}^{- 2 t} - 3 {e}^{t} + 5 y = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}} + 5 y = 3 {e}^{t}$ ..... [A]

This is now a second order linear 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.

${m}^{2} - m + 5 = 0$

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

$m = \frac{1}{2} \left(1 \pm \sqrt{19} i\right)$

Thus the Homogeneous equation:

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}} + 5 y = 0$

has the solution:

$y = {e}^{\frac{1}{2} t} \left(A \cos \left(\frac{1}{2} \sqrt{19} t\right) + B \cos \left(\frac{1}{2} \sqrt{19} t\right)\right)$

We now seek a Particular Solution of [A], which will be of the form:

$y = A {e}^{t}$

Differentiating wrt $t$ we get:

$y ' = A {e}^{t}$, and, >$y ' ' = A {e}^{t}$

Substituting into the DE [A} we get:

$A {e}^{t} - A {e}^{t} + 5 A {e}^{t} = 3 {e}^{t}$

Equating coefficients we have:

$A - A + 5 A = 3 \implies A = \frac{3}{5}$

Hence, the GS of [A] is the combination of the homogeneous solution and the particular solution, thus:

$y = {e}^{\frac{1}{2} t} \left(A \cos \left(\frac{1}{2} \sqrt{19} t\right) + B \cos \left(\frac{1}{2} \sqrt{19} t\right)\right) + \frac{3}{5} {e}^{t}$

Now we initially used a change of variable:

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

So restoring this change of variable we get:

$y = {e}^{\frac{1}{2} \ln x} \left(A \cos \left(\frac{1}{2} \sqrt{19} \ln x\right) + B \cos \left(\frac{1}{2} \sqrt{19} \ln x\right)\right) + \frac{3}{5} {e}^{\ln} x$

$\therefore y = {e}^{\ln \sqrt{x}} \left(A \cos \left(\frac{1}{2} \sqrt{19} \ln x\right) + B \cos \left(\frac{1}{2} \sqrt{19} \ln x\right)\right) + \frac{3}{5} x$

$\therefore y = \sqrt{x} \left(A \cos \left(\frac{1}{2} \sqrt{19} \ln x\right) + B \cos \left(\frac{1}{2} \sqrt{19} \ln x\right)\right) + \frac{3}{5} x$

Which is the General Solution