# What is the general solution of the differential equation  x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x ?

Oct 21, 2017

$y = A {x}^{2 - \sqrt{3}} + B {x}^{2 + \sqrt{3}} + \frac{5 \sin \ln x}{61 x} + \frac{6 \cos \ln x}{61 x}$

#### Explanation:

We have:

${x}^{2} \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} + y = \sin \frac{\log x}{x}$ ..... [A]

This is a Euler-Cauchy Equation (the power of $x$ is the same as the degree of the differential in every occurrence of their product) 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:

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

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - \frac{\mathrm{dy}}{\mathrm{dt}} - 3 \frac{\mathrm{dy}}{\mathrm{dt}} + y = {e}^{- t} \sin t$

$\therefore \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} - 4 \frac{\mathrm{dy}}{\mathrm{dt}} + y = {e}^{- t} \sin t$ ..... [B]

This is now a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, ${y}_{p}$ of the non-homogeneous equation.

Complementary Function

The associated homogeneous equation is:

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

Which has the Auxiliary Equation:

${m}^{2} - 4 m + 1 = 0$

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

$m = 2 \pm \sqrt{3}$

Thus the Homogeneous equation [C] has the solution:

${y}_{c} = A {e}^{\left(2 - \sqrt{3}\right) t} + B {e}^{\left(2 + \sqrt{3}\right) t}$

Particular Solution

With this particular equation [B], a probably solution is of the form:

$y = a {e}^{- t} \sin t + b {e}^{- t} \cos t$
$\setminus \setminus = {e}^{- t} \left(a \sin t + b \cos t\right)$

Where $a$ and $b$ are constants to be determined. Let us assume the above solution works, in which case be differentiating wrt $x$ we have:

$y ' \setminus \setminus = {e}^{- t} \left(a \cos t - b \sin t\right) - {e}^{- t} \left(a \sin t + b \cos t\right)$
$\setminus \setminus \setminus \setminus \setminus = {e}^{- t} \left(\left(a - b\right) \cos t - \left(a + b\right) \sin t\right)$

$y ' ' = {e}^{- t} \left(- \left(a - b\right) \sin t - \left(a + b\right) \cos t\right) - {e}^{- t} \left(\left(a - b\right) \cos t - \left(a + b\right) \sin t\right)$
$\setminus \setminus \setminus \setminus \setminus = {e}^{- t} \left(2 b \sin t - 2 a \cos t\right)$

Substituting into the Differential Equation $\left[B\right]$ we get:

${e}^{- t} \left(2 b \sin t - 2 a \cos t\right) - 4 {e}^{- t} \left(\left(a - b\right) \cos t - \left(a + b\right) \sin t\right) + {e}^{- t} \left(a \sin t + b \cos t\right) = {e}^{- t} \sin t$

Equating coefficients of $\cos \left(x\right)$ and $\sin \left(x\right)$ we get:

$\cos \left(x\right) : - 2 a - 4 \left(a - b\right) + b = 0$
$\sin \left(x\right) : 2 b + 4 \left(a + b\right) + a = 1$

Solving simultaneously we get:

$a = \frac{5}{61}$ and $b = \frac{6}{61}$

And so we form the Particular solution:

${y}_{p} = \frac{5}{61} {e}^{- t} \sin t + \frac{6}{61} {e}^{- t} \cos t$

General Solution

Which then leads to the GS of [B}

$y \left(t\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A {e}^{\left(2 - \sqrt{3}\right) t} + B {e}^{\left(2 + \sqrt{3}\right) t} + \frac{5}{61} {e}^{- t} \sin t + \frac{6}{61} {e}^{- t} \cos t$

Now we initially used a change of variable:

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

So restoring this change of variable we get:

$y = A {e}^{\left(2 - \sqrt{3}\right) \ln x} + B {e}^{\left(2 + \sqrt{3}\right) \ln x} + \frac{5}{61} {e}^{- \ln x} \sin \ln x + \frac{6}{61} {e}^{- \ln x} \cos \ln x$

$\therefore y = A {x}^{2 - \sqrt{3}} + B {x}^{2 + \sqrt{3}} + \frac{5}{61} {e}^{\ln \left(\frac{1}{x}\right)} \sin \ln x + \frac{6}{61} {e}^{\ln \left(\frac{1}{x}\right)} \cos \ln x$

$\setminus \setminus \setminus \setminus \setminus \setminus = A {x}^{2 - \sqrt{3}} + B {x}^{2 + \sqrt{3}} + \frac{5}{61} \left(\frac{1}{x}\right) \sin \ln x + \frac{6}{61} \left(\frac{1}{x}\right) \cos \ln x$

$\setminus \setminus \setminus \setminus \setminus \setminus = A {x}^{2 - \sqrt{3}} + B {x}^{2 + \sqrt{3}} + \frac{5 \sin \ln x}{61 x} + \frac{6 \cos \ln x}{61 x}$

Which is the General Solution of [A].