Question

In each of the following examples verify that the given function is a solution of the corresponding differential equation.
(i) $y=c_1\sin x+c_2\cos x;\frac{d^{2}y}{d{x}^2}+y=0$
(ii) $y\sec x=\tan x+c;\frac{dy}{dx}+y\tan x=\sec x$

Answer 1

$\left(i\right)$ given differential equation.

$\frac{d^{2}y}{d{x}^2}+y=\mathrm{0.....}\left(1\right)$

corresponding solution of above differential equation is

$y=c_1\sin x+c_2\cos x....\left(2\right)$

if $y$ is a solution it must satisfy given differential equation.

differentiate equation (2) with respect to x

$\frac{dy}{dx}=c_1\cos x-c_2\sin x$

again differentiate above equation with respect to x we get.

$\frac{d^{2}y}{d{x}^2}=-c_1\sin x-c_2\cos x$

or

$\frac{d^{2}y}{d{x}^2}+c_1\sin x+c_2\cos x=\mathrm{0.....}\left(3\right)$

from equation (2) we have $y=c_1\sin x+c_2\cos x$

Hence,

$\frac{d^{2}y}{d{x}^2}+y=0$

Hence, $y$ is solution of given differential equation.

$\left(ii\right)$ given differential equation.

$\frac{dy}{dx}+y\tan x=\sec x...\left(4\right)$

corresponding solution of above differential equation is

$y\sec x=\tan x+c...\left(5\right)$

if $y$ is a solution it must satisfy given differential equation.

differentiate equation (4) with respect to x

$\sec x\frac{dy}{dx}+y\tan x\sec x=se{c}^{2}x$

or

$\frac{dy}{dx}+y\tan x=\sec x$

Hence, $y$ is solution of given differential equation (4).