Question

Let $y=y\left(x\right)$ be the solution of the differential equation $\cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3\left)\right)dx,0\le x\le \frac{\mathrm{\pi }}{2},y\left(0\right)=0.$ Then, $y\left(\frac{\mathrm{\pi }}{3}\right)$ is equal to:

• A. $2{\log }_e\left(\frac{2\sqrt{3}+10}{11}\right)$
• B. $2{\log }_e\left(\frac{\sqrt{3}+7}{2}\right)$
• C. $2{\log }_e\left(\frac{3\sqrt{3}-8}{4}\right)$
• D. $2{\log }_e\left(\frac{2\sqrt{3}+9}{6}\right)$

Answer 1

The correct option is A

$2{\log }_e\left(\frac{2\sqrt{3}+10}{11}\right)$

Explanation for the correct option:

Step-1: convert in linear differential equation form and find $\mathbf{I}\mathbf{.}\mathbf{F}$:

Given,

$\cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3\left)\right)dx,y\left(0\right)=0.$

On solving, we get:

$$\begin{gathered} \cos x(3\sin x+\cos x+3)dy=(1+y\sin x(3\sin x+\cos x+3\left)\right)dx \\ \Rightarrow \frac{dy}{dx}-y\left(\tan x\right)=\frac{1}{\cos x(3\sin x+\cos x+3)} \end{gathered}$$

$\begin{array}{rcl}I.F.& =& e^{\int -\tan xdx}\\ & =& e^{\ln \left|\cos x\right|}\\ & =& \left|\cos x\right|\\ & =& \cos x\left[\because 0\le x\le \frac{\pi }{2}\right]\end{array}$

Step-2: Solution of linear differential equation:

Solving the differential equation:

$$\begin{gathered} y\left(\cos x\right)=\int \left(\cos x\right).\frac{1}{\cos x(3\sin x+\cos x+3)}dx \\ \Rightarrow y(\cos x)=\int \frac{1}{(3\sin x+\cos x+3)}dx \\ \Rightarrow y(\cos x)=\int \frac{dx}{3\left({\displaystyle \frac{2\tan \left({\displaystyle \frac{x}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{x}{2}}\right)}}\right)+\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}\right)+3} \\ \Rightarrow y(\cos x)=\int \frac{se{c}^2{\displaystyle \frac{x}{2}}}{2{\tan }^2{\displaystyle \frac{x}{2}}+6\tan {\displaystyle \frac{x}{2}}+4}dx \end{gathered}$$

Let,

Step-3 : Integration and final solution of differential equation:

$$\begin{gathered} I_1=\int \frac{se{c}^2{\displaystyle \frac{x}{2}}}{2{\tan }^2{\displaystyle \frac{x}{2}}+6\tan {\displaystyle \frac{x}{2}}+4}dx \\ \text{Substituting}\tan \frac{x}{2}=t, \\ \Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt \end{gathered}$$

$\begin{array}{rcl}\therefore \int \frac{dt}{t^3+3t+2}& =& \frac{dt}{(t+2)(t+1)}\\ & =& \int \left(\frac{1}{t+1}-\frac{1}{t+2}\right)dt\\ & =& \ln \left|\left(\frac{t+1}{t+2}\right)\right|\\ & =& \ln \left|\frac{\tan {\displaystyle \frac{x}{2}}+1}{\tan {\displaystyle \frac{x}{2}}+2}\right|+C\end{array}$

Finding the solution of the differential equation:

$$\begin{gathered} y\left(\cos x\right)=\ln \left|\frac{\tan {\displaystyle \frac{x}{2}}+1}{\tan {\displaystyle \frac{x}{2}}+2}\right|+C \\ \Rightarrow y\left(\cos x\right)=\ln \left|\frac{\tan {\displaystyle \frac{x}{2}}+1}{\tan {\displaystyle \frac{x}{2}}+2}\right|+Cfor0\le x<\frac{\mathrm{\pi }}{2} \end{gathered}$$

Now, it is given that $y\left(0\right)=0$.

Therefore,

$$\begin{gathered} \Rightarrow 0=\ln \left(\frac{1}{2}\right)+C \\ \Rightarrow C=\ln \left(2\right) \\ \Rightarrow y\left(\cos x\right)=\ln \left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{2+\tan {\displaystyle \frac{x}{2}}}\right)+\ln 2 \\ Forx=\frac{\mathrm{\pi }}{3} \\ y\left(\frac{1}{2}\right)=\ln \left(\frac{1+{\displaystyle \frac{1}{\sqrt{3}}}}{2+\frac{1}{\sqrt{3}}}\right)+\ln 2\left[\because cos\frac{\pi }{3}=\frac{1}{2}\right] \\ \Rightarrow y=2\ln \left(\frac{2\sqrt{3}+10}{11}\right)\left[\because \mathrm{Rationlize}\mathrm{the}\mathrm{denominator}\right] \end{gathered}$$

Hence, the correct answer is option (A).