What is a second solution to the Differential Equation  x^2y'' -3xy'+5y=0 ?

We are given that the answer does not contain cosine functions.

Jun 16, 2017

See below.

Explanation:

Assuming that the differential equation reads

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

The differential equation

${x}^{2} y ' ' - 3 x y ' + 5 y = 0$ is a linear homogeneous differential equation.

Proposing

$y = {c}_{0} {x}^{\alpha}$ and substituting

$\left(\alpha \left(\alpha - 4\right) + 5\right) {c}_{0} {x}^{\alpha} = 0$ and solving for $\alpha$

$\alpha \left(\alpha - 4\right) + 5 = \left(\alpha - 2 + i\right) \left(\alpha - 2 - i\right)$

then

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

but $x = {e}^{{\log}_{e} x}$ and ${x}^{\pm i} = {e}^{\pm i {\log}_{e} x}$

now according to

${e}^{i z} = \cos z + i \sin z$ we have the general solution.

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

Jun 21, 2017

Second solution is:

$y = B {x}^{2} \sin \left(\ln x\right)$

Explanation:

If we assume the a corrected equation:

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

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

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

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - 4 \frac{\mathrm{dy}}{\mathrm{dt}} + 5 y = 0$ ..... [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.

${m}^{2} - 4 m + 5 = 0$

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

$m = 2 \pm i$

Thus the Homogeneous equation [B]:

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - 4 \frac{\mathrm{dy}}{\mathrm{dt}} + 5 y = 0$

has the solution:

$y = {e}^{2 t} \left(A \cos t + B \sin t\right)$

Now we initially used a change of variable:

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

So restoring this change of variable we get:

$y = {\left(x\right)}^{2} \left(A \cos \left(\ln x\right) + B \sin \left(\ln x\right)\right)$

$\therefore y = {x}^{2} \left(A \cos \left(\ln x\right) + B \sin \left(\ln x\right)\right)$

$\therefore y = A {x}^{2} \cos \left(\ln x\right) + B {x}^{2} \sin \left(\ln x\right)$

Which is the General Solution.

From the given answer we can clearly infer that for the first part of the solution $A = 0$, but we cannot make any further assumptions about the constant $B$ in the second solution