Question

Check if the given functions belong to the solution space of some second
order differential equation
f (x) = 9 cos (2x) 9 (x) = 2cos? (x) – 2sin? (x)​

Answer 1

$\textbf{Given:}$

$f(x)=9\,cos2x$

$g(x)=2\,cosx-2\,sinx$

$\textbf{To find:}$

$\text{whether the given functions belong to the solution space}$

$\text{of some second order differential equation}$

$\textbf{Solution:}$

$\text{Consider,}$

$f(x)=9\,cos2x$

$\text{Differentiate with respect to x}$

$\dfrac{df}{dx}=9(-2\,sin2x)$

$\text{Differentiate again with respect to x}$

$\dfrac{d^2f}{dx^2}=9(-2^2\,cos2x)$

$\dfrac{d^2f}{dx^2}=-4(9\,cos2x)$

$\dfrac{d^2f}{dx^2}=-4\,f$

$\implies\dfrac{d^2f}{dx^2}+4\,f=0$

$\therefore\bf\,f(x)=9\,cos2x\;\textbf{is the solution of the second order differential equation}\\\bf\dfrac{d^2f}{dx^2}+4\,f=0$

$\text{Consider,}$

$g(x)=2\,cosx-2\,sinx$

$\text{Differentiate with respect to x}$

$\dfrac{dg}{dx}=2(-sinx)-2(cosx)$

$\text{Differentiate again with respect to x}$

$\dfrac{d^2g}{dx^2}=2(-cosx)-2(-sinx)$

$\dfrac{d^2g}{dx^2}=-2\,cosx+2\,sinx$

$\dfrac{d^2g}{dx^2}=-(2\,cosx-2\,sinx)$

$\dfrac{d^2g}{dx^2}=-g$

$\implies\dfrac{d^2g}{dx^2}+g=0$

$\therefore\bf\,g(x)=2\,cosx-2\,sinx\;\textbf{is the solution of the second order differential equation}\\\bf\dfrac{d^2g}{dx^2}+g=0$