Question

The solution of$dy=\cos x(2-y\cos ecx)dx$, where$y=2$, where$x=\frac{\pi }{2}$is

• A. $y=\sin x+\cos ecx$
• B. $y=\tan \left(\frac{x}{2}\right)+cot\left(\frac{x}{2}\right)$
• C. $y=\left(\frac{1}{\sqrt{2}}\right)sec\left(\frac{x}{\sqrt{2}}\right)+\sqrt{2}\cos \left(\frac{x}{2}\right)$
• D. None of the above

Answer 1

The correct option is D

None of the above

Explanation of the correct option.

Step-1 Converting in linear differential equation:

Given:$dy=\cos x(2-y\cos ecx)dx$

We can also write as,

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\cos \left(x\right)(2-y\csc \left(x\right)) \\ \Rightarrow \frac{dy}{dx}=2\cos \left(x\right)-y\frac{\cos \left(x\right)}{\sin \left(x\right)} \\ \Rightarrow \frac{dy}{dx}=2\cos \left(x\right)-y\cot \left(x\right) \\ \Rightarrow \frac{dy}{dx}+y\cot \left(x\right)=2\cos \left(x\right) \end{gathered}$$

Step-2 : Finding integrating factor:

Since the solution of first-order differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is given by, $\mathbf{y}\left(I.F.\right)\mathbf{=}\mathbf{\int }\left(I.F.\right)\mathbf{}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{dx}$

We know that, $I.F.$ is calculated as $\mathbf{I}\mathbf{.}\mathbf{F}\mathbf{.}\mathbf{=}{\mathbf{e}}^{\mathbf{\int }\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{dx}}$

So

,$\begin{array}{rcl}I.F.& =& e^{\int \cot \left(x\right)dx}\left[\int \cot \left(x\right)\mathrm{dx}=\ln \left(\sin \left(x\right)\right)\right]\\ & =& e^{\ln \left(\sin \left(x\right)\right)}\left[e^{\ln \left(x\right)}=x\right]\\ & =& \sin \left(x\right)\end{array}$

Step-3: Solution of differential equation:

Therefore, the general solution is

$$\begin{gathered} y\sin \left(x\right)=\int 2\cos \left(x\right)\sin \left(x\right)dx+C \\ \Rightarrow y\sin \left(x\right)=\int \sin \left(2x\right)dx+C \end{gathered}$$

Now compute$\int \sin \left(2x\right)dx$

Let, $2x=u$

$$\begin{gathered} \Rightarrow 2\left(1\right)=\frac{du}{dx} \\ \Rightarrow dx=\frac{du}{2} \end{gathered}$$

$\begin{array}{rcl}\int \sin \left(2x\right)dx& =& \int \sin \left(u\right)\frac{du}{2}\\ & =& \frac{1}{2}(-\cos \left(u\right))+c\\ & =& -\frac{\cos \left(2x\right)}{2}+c\end{array}$
Now,

$\Rightarrow y\sin \left(x\right)=-\frac{\cos \left(2x\right)}{2}+C.....................\left(1\right)$

at $y=2$ and $x=\frac{\pi }{2}$

$$\begin{gathered} \Rightarrow 2\sin \left(\frac{\mathrm{\pi }}{2}\right)=-\frac{\cos \left(2{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{2}+C\left[\because \sin \left(\frac{\pi }{2}\right)=1,\cos \left(\pi \right)=-1\right] \\ \Rightarrow 2=-\frac{(-1)}{2}+C \\ \Rightarrow 2=\frac{1}{2}+C \\ \Rightarrow C=\frac{3}{2} \end{gathered}$$
From equation $\left(1\right)$,

$\therefore y\sin \left(x\right)=-\frac{\cos \left(2x\right)}{2}+\frac{3}{2}$

Hence, option (D) is the correct answer.