Q28RP

Question

Find a general solution to the given differential equation.

$y'' = 5 x^{- 1} y' - 13 x^{- 2} y, x > 0$

Step-by-Step Solution

Verified

Answer

The general solution to the given differential equation is;

$y = c_{1} x^{3} \cos (2 lnx) + c_{2} x^{3} \sin (2 lnx)$

1Write the auxiliary equation of the given differential equation

The given differential equation is,

$\begin{array}{rcl} y'' & = & 5 x^{- 1} y' - 13 x^{- 2} y \\ y'' - 5 x^{- 1} y' + 13 x^{- 2} y & = & 0 \\ x^{2} y'' - 5 xy' + 13 y & = & 0 . . . . . . . (1) \end{array}$

Let,

$\begin{array}{rcl} x & = & e^{t} \\ dx & = & e^{t} dt \\ \frac{dt}{dx} & = & e^{- t} \\ y' & = & \frac{dy}{dx} = \frac{dy}{dt} e^{- t} \\ y'' & = & \frac{d^{2} y}{{dx}^{2}} = e^{- t} (\frac{dy}{dt} e^{- t} (- 1) + \frac{d^{2} y}{{dt}^{2}} e^{- t}) = e^{- 2 t} (\frac{d^{2} y}{{dt}^{2}} - \frac{dy}{dt}) \end{array}$

Substitute the value of $x, y'$ and $y''$ in the equation (1),

$\begin{array}{rcl} x^{2} y'' - 5 xy' + 13 y & = & 0 \\ e^{2 t} [e^{- 2 t} (\frac{d^{2} y}{{dt}^{2}} - \frac{dy}{dt})] - 5 e^{t} [\frac{dy}{dt} e^{- t}] + 13 y & = & 0 \\ \frac{d^{2} y}{{dt}^{2}} - \frac{dy}{dt} - 5 \frac{dy}{dt} + 13 y & = & 0 \\ y'' - 6 y' + 13 y & = & 0 . . . . . . (2) \end{array}$

The auxiliary equation for the above equation, $m^{2} - 6 m + 13 = 0 .$

2Find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{rcl} m^{2} - 6 m + 13 & = & 0 \\ m & = & \frac{6 \pm \sqrt{36 - 52}}{2} \\ m & = & \frac{6 \pm \sqrt{- 16}}{2} \\ m & = & 3 \pm 2 i \end{array}$