Question

Obtain the differential equation by eliminating the arbitrary constants from the given equation, y = A cos (log x) + B sin (log x)

Answer 1

To obtain the differential equation by eliminating the arbitrary constants from the given equation, y = A cos (log x) + B sin (log x)
differentiate the given equation twice

$y = A \: cos \: (log \: x) + B \: sin \: (log \: x) ...eq1\\ \\ \frac{dy}{dx} = - A \: sin \: (log \: x)( \frac{1}{x} ) + B \: cos \: (log \: x)( \frac{1}{x} )...eq2 \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = - A \: sin \: ( log \: x)( \frac{ - 1}{ {x}^{2} } ) - \frac{A}{x} \: cos \: (log \: x)( \frac{1}{x} ) + B \: cos(log \: x)( - \frac{1}{ {x}^{2} } ) + \frac{B}{x} ( - sin \: (log \: x) (\frac{1}{x} ) \\ \\ = \frac{a}{ {x}^{2} } sin \: (log \: x) - \frac{A}{ {x}^{2} } cos \: (log \: x) - \frac{B}{ {x}^{2} } cos \: (log \: x) - \frac{B}{ {x}^{2} } sin \: (log \: x) \\ \\ = - \frac{1}{ {x}^{2} } (A \: cos \: (log \: x) + B \: sin \: (log \: x) + \frac{1}{x} ( \frac{A}{x} \: sin \: (log \: x) - \frac{B}{x} cos \: (log \: x)$
substitute the values from eq 1 and 2

$\frac{ {d}^{2} y}{ {dx}^{2} } = - \frac{1}{ {x}^{2} } (A \: cos \: (log \: x) + B \: sin \: (log \: x) + \frac{1}{x} ( \frac{A}{x} \: sin \: (log \: x) - \frac{B}{x} cos \: (log \: x) \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = - \frac{1}{ {x}^{2} } y - \frac{1}{x} \frac{dy}{dx} \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } + \frac{y}{ {x}^{2} } + \frac{1}{x} \frac{dy}{dx} = 0 \\ \\ {x}^{2} \frac{ {d}^{2} y}{ {dx}^{2} } + x \frac{dy}{dx} + y = 0$
is the differential equation.