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

Jun 10, 2017

see Below

#### Explanation:

There is a certain way to find the general solution if a one solution is already given. It is said that if given solution ${y}_{1}$, the second one is given as ${y}_{2} = v \left(x\right) {y}_{1}$. Where in this problem, ${y}_{1} = \frac{1}{x}$

If ${y}_{2} = v \left(x\right) \frac{1}{x}$, then the derivatives are:
$y {'}_{2} = v \left(- \frac{1}{{x}^{2}}\right) + v ' \left(\frac{1}{x}\right)$

and

$y ' {'}_{2} = v \left(\frac{2}{x} ^ 3\right) + v ' \left(- \frac{2}{x} ^ 2\right) + v ' ' \left(\frac{1}{x}\right)$

Substitute this values in the original equation:

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

And:

${x}^{2} \left(v \left(\frac{2}{x} ^ 3\right) + v ' \left(- \frac{2}{x} ^ 2\right) + v ' ' \left(\frac{1}{x}\right)\right) - x \left(v \left(- \frac{1}{{x}^{2}}\right) + v ' \left(\frac{1}{x}\right)\right) - 3 \left(v \left(\frac{1}{x}\right)\right) = 0$

Although this is suprisingly long, this actually simplifies to:

$v ' ' - \frac{3}{x} v ' = 0$

Substitute $w = v '$ so that is easy to solve as first order differential equation.

$w ' - \frac{3}{x} w = 0$

The integrating factor is $\frac{1}{x} ^ 3$(Given by ${e}^{\int \frac{3}{x} \mathrm{dx}}$).

$\frac{1}{x} ^ 3 w ' - \frac{3}{x} ^ 4 w = 0$
$\left(\frac{1}{x} ^ 3 w\right) ' = 0$

Integrate both sides:

$\frac{1}{x} ^ 3 w = {C}_{1}$

Undo substitution:

$v ' \frac{1}{x} ^ 3 = {C}_{1}$
$v ' = {C}_{1} {x}^{3}$

Integrate both sides again:

$v = {C}_{1} / 4 {x}^{4} + {C}_{2}$

Now the general solution is given as:
${y}_{2} = v \left(x\right) {y}_{1}$
${y}_{2} = \left({C}_{1} / 4 {x}^{4} + {C}_{2}\right) \left(\frac{1}{x}\right)$
${y}_{2} = \left({C}_{1} \left({x}^{3} / 4\right) + {C}_{2} \left(\frac{1}{x}\right)\right)$

Now the first solution is given when ${C}_{1} = 0$, then the second solution can be when ${C}_{2} = 0$ which is ${x}^{3} / 4$ otherwise there are infinite solutionsby just filling the constants.

To check:
$y = {x}^{3} / 4$
$y ' = \frac{3}{4} {x}^{2}$
$y ' ' = \frac{3}{2} x$

Substitute:
${x}^{2} \left(\frac{3}{2} x\right) - x \left(\frac{3}{4} {x}^{2}\right) - 3 \left({x}^{3} / 4\right) = 0$
$\frac{3}{2} {x}^{3} - \frac{3}{4} {x}^{3} - \frac{3}{4} {x}^{3} = 0$
$6 {x}^{3} - 3 {x}^{3} - 3 {x}^{3} = 0$
$0 = 0$

Which is correct.

Jun 17, 2017

$y = \frac{A}{x} + B {x}^{3}$

#### Explanation:

We have:

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

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

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

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - 2 \frac{\mathrm{dy}}{\mathrm{dt}} - 3 y = 0$ ..... [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.

${m}^{2} - 2 m - 3 = 0$

We can solve this quadratic equation, and we get two real and distinct solutions:

$\left(m + 1\right) \left(m - 3\right) = 0 \implies m = - 1 , 3$

Thus the Homogeneous equation [B]:

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - 2 \frac{\mathrm{dy}}{\mathrm{dt}} - 3 y = 0$

has the solution:

$y = A {e}^{- t} + B {e}^{3 t}$

Now we initially used a change of variable:

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

So restoring this change of variable we get:

$y = A {e}^{- \ln x} + B {e}^{3 \ln x}$

$\therefore y = A {e}^{\ln {x}^{- 1}} + B {e}^{\ln {x}^{3}}$

$\therefore y = A {x}^{- 1} + B {x}^{3}$

$\therefore y = \frac{A}{x} + B {x}^{3}$

Which is the General Solution