Question

If $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ is the solution of the differential equation, $\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{ytan}\mathrm{x}=\sin \mathrm{x},\mathrm{y}\left(\frac{\mathrm{\pi }}{3}\right)=0$, then the maximum value of the function $\mathrm{y}\left(\mathrm{x}\right)$ over $\mathrm{R}$ is equal to

• A. $8$
• B. $\frac{1}{2}$
• C. $-\frac{15}{4}$
• D. $\frac{1}{8}$

Answer 1

The correct option is D

$\frac{1}{8}$

Explanation for the correct option:

Finding the maximum value of the function $\mathrm{y}\left(\mathrm{x}\right)$ over $\mathrm{R}$:

The given differential equation is $\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{ytan}\mathrm{x}=\sin \mathrm{x},\mathrm{y}\left(\frac{\mathrm{\pi }}{3}\right)=0$

$\begin{array}{l}\mathrm{I}.\mathrm{F}.={\mathrm{e}}^{\int 2\mathrm{tanxdx}}={\mathrm{e}}^{2\mathrm{lnsecx}}\\ \mathrm{I}.\mathrm{F}.={\sec }^2\mathrm{x}\end{array}$

The differential equation is

$$\begin{gathered} \mathrm{y}·\mathrm{I}.\mathrm{F}=\int \mathrm{Q}\left(\mathrm{x}\right)\mathrm{I}.\mathrm{Fdx} \\ \begin{array}{c}\Rightarrow \mathrm{y}.\left({\sec }^2\mathrm{x}\right)=\int \mathrm{sinx}.{\sec }^2\mathrm{xdx}\\ =\int \mathrm{secxtanxdx}\\ =\mathrm{secx}+\mathrm{C}...\left(1\right)\end{array} \end{gathered}$$

Put $\mathrm{x}=\frac{\mathrm{\pi }}{3};\mathrm{y}=0$

$$\begin{gathered} \left(0\right)se{c}^2\left(\frac{\mathrm{\pi }}{3}\right)=sec\left(\frac{\mathrm{\pi }}{3}\right)+C \\ \Rightarrow C=-2 \end{gathered}$$

Substitute the value of $\mathrm{C}$ in the equation $\left(1\right)$.

$\begin{array}{rcl}\mathrm{y}& =& \frac{\mathrm{secx}-2}{{\sec }^2\mathrm{x}}\\ & =& \mathrm{cosx}-2{\cos }^2\mathrm{x}\left[\because \mathrm{cos\theta }=\frac{1}{\mathrm{sec\theta }}\right]\end{array}$

Assume that $\cos \mathrm{x}=\mathrm{t}$

$\mathrm{y}=\mathrm{t}-2{\mathrm{t}}^2...\left(2\right)$

Differentiating with respect to $\mathrm{y}.$

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dt}}=1-4\mathrm{t} \\ \Rightarrow 1-4\mathrm{t}=0 \\ \begin{array}{rcl}& \Rightarrow & \mathrm{t}=\frac{1}{4}...\left(3\right)\end{array} \end{gathered}$$

Substitute $\left(3\right)$ in $\left(2\right)$,

$\begin{array}{c}y=\frac{1}{4}-2{\left(\frac{1}{4}\right)}^2\\ =\frac{1}{4}-2\left(\frac{1}{16}\right)\\ =\frac{1}{4}-\frac{1}{8}\\ =\frac{2-1}{8}\\ =\frac{1}{8}\end{array}$

Hence, the correct option is D.