Question

In each of the following examples verify that the given function is a solution of the corresponding differential equation.
(i) $y=c_{1}\sin x+c_{2}\cos x$; $\dfrac{d^{2}y}{dx^{2}} +y=0$.
(ii) $y\sec x=\tan x+c$; $\dfrac{dy}{dx} +y\tan x=\sec x$.

Answer 1

Hint: In this question it is given that whether the functions $y=c_{1}\sin x+c_{2}\cos x$ and $y\sec x=\tan x+c$ satisfies the given differential equations or not, so for finding solution we have to take the given function and have to differentiate both side with respect to ‘x’ and if the required differential equation is of second order then we have to differentiate that function twice.

Complete step-by-step solution:
(i) Given function,
$y=c_{1}\sin x+c_{2}\cos x$ …………….equation (1)
Now differentiating both of the side of the above function w.r.t ‘x’ we get,
${}\dfrac{dy}{dx} =\dfrac{d}{dx} \left( c_{1}\sin x+c_{2}\cos x\right)$
$\Rightarrow {}\dfrac{dy}{dx} =\dfrac{d}{dx} \left( c_{1}\sin x\right) +\dfrac{d}{dx} \left( c_{2}\cos x\right)$
$\Rightarrow {}\dfrac{dy}{dx} =c_{1}\dfrac{d}{dx} \left( \sin x\right) +c_{2}\dfrac{d}{dx} \left( \cos x\right)$
$\Rightarrow {}\dfrac{dy}{dx} =c_{1}\cos x+c_{2}\left( -\sin x\right)$
[ since we know that $\dfrac{d}{dx} \left( \sin x\right) =\cos x$ and $\dfrac{d}{dx} \left( \cos x\right) =-\sin x$]
$\Rightarrow \dfrac{dy}{dx} =c_{1}\cos x-c_{2}\sin x$ ………...equation (2)
Now again differentiating equation (2) w.r.t ‘x’ we get,
$\dfrac{d^{2}y}{dx^{2}} =\dfrac{d}{dx} \left( c_{1}\cos x-c_{2}\sin x\right)$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{d}{dx} \left( c_{1}\cos x\right) -\dfrac{d}{dx} \left( c_{2}\sin x\right)$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =c_{1}\dfrac{d}{dx} \left( \cos x\right) -c_{2}\dfrac{d}{dx} \left( \sin x\right)$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =c_{1}\left( -\sin x\right) -c_{2}\left( \cos x\right)$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =-c_{1}\sin x-c_{2}\cos x$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =-\left( c_{1}\sin x+c_{2}\cos x\right)$
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =-y$ [by using equation (1)]
$\Rightarrow \dfrac{d^{2}y}{dx^{2}} +y=0$
Hence, we can say that the given function $y=c_{1}\sin x+c_{2}\cos x$ is the solution of the given differential equation $\dfrac{d^{2}y}{dx^{2}} +y=0$.

(ii) Given function,
$y\sec x=\tan x+c$ …………..equation (3)
Now differentiating both of the side of the above function w.r.t ‘x’ we get,
$\dfrac{d}{dx} \left( y\sec x\right) =\dfrac{d}{dx} \left( \tan x+c\right)$
Since as we know that, $\dfrac{d}{dx} \left( uv\right) =u\dfrac{dv}{dx} +v\dfrac{du}{dx}$, where u and v are the function of x, so by using the above formula,we can write,
$y\dfrac{d}{dx} \left( \sec x\right) +\sec x\dfrac{dy}{dx} =\dfrac{d}{dx} \left( \tan x\right) +\dfrac{d}{dx} \left( c\right)$
$\Rightarrow y\left( \sec x\tan x\right) +\sec x\dfrac{dy}{dx} =\sec^{2} x$
Now taking $\sec x$ common from both side of the equation and cancelling with each other, we get,
$y\tan x+\dfrac{dy}{dx} =\sec x$
$\Rightarrow \dfrac{dy}{dx} +y\tan x=\sec x$
Hence, we can say that the given function $y\sec x=\tan x+c$ is the solution of the given differential equation $\dfrac{dy}{dx} +y\tan x=\sec x$.

Note: While solving this type of question you have to know the procedure that you have to transform a normal equation into a differential equation, and if the obtained differential equation matches with the given differential equation then we are able to say that the given function is the solution of obtained differential equations.