Question

If $y = y\left( x \right)$ is the solution of differential equation, $ \dfrac{{dy}}{{dx}} + 2y\tan x = {{ sin}}x,$ $ y \left( {\dfrac{\pi }{{ 3}} } \right) = 0,$ Then the maximum value of $y\left( x \right)$ over$ R $is equal to
A. $8$
B. $\dfrac{1}{2}$
C. $ - \dfrac{{ 15}}{4}$
D. $\dfrac{1}{8}$

Answer 1

Hint: We will find the Integrating factor (IF) for the given linear first-order differential equation and evaluate the differentiation of given function. Then substitute the value of given value of $x = \dfrac{\pi }{{ 3}}$ then $y = 0$ to obtain the maximum value of $y\left( x \right)$.

Formula Used:
The formula for the integrating factor IF is ${e^{\int {Pdx} }}$ for the linear first-order differential equation written in standard form: $\dfrac{{dy}}{{dx}} + p\left( x \right)y = q\left( x \right).$Then the solution of such equation is given by ${{IF}} \times \dfrac{{dy}}{{dx}} + {{ IF}} \times P\left( x \right)y = {{ IF}} \times Q\left( x \right)$.

Complete step by step solution:
Evaluate the integrating factor of given linear first-order differential equation
$ \dfrac{{dy}}{{dx}} + 2y{{ tan}}x = {{ sin}}x$
The integrating factor (IF) will be,
${{IF }} = {e^{\int {2\tan x dx} }}$
$IF = {e^{\log \left( {{{\sec }^2}x} \right) }}$
$IF = {{ se}}{{{c}}^2}\;x$
Now multiply the IF with each term of given linear first-order differential equation, we get
${\sec ^2}\;x\dfrac{{dy}}{{dx}} + 2y{{ tan}}x\left( {{{\sec }^2}\;x} \right) = {{ sin}}x\left( {{{\sec }^2}\;x} \right)$
Further evaluating the equation, we get
${{ se}}{{{c}}^2}\;x\left( {dy} \right) + 2y{{ tan}}x\left( {{{\sec }^2}\;x} \right)dx = {{ sin}}x\left( {{{\sec }^2}\;x} \right)dx$
Now integrate the above equation,
$y{{ se}}{{{c}}^2}x = \smallint \tan x{{ sec}}x dx = {{ sec}}x + c$
Now put $x = \dfrac{\pi }{3} $, $y = 0,$ $c = - 2$ ,we get
$y = \cos x-2{\cos ^2}x$
Taking $ - 2$ common from right-hand side, we get
$y = -2 \left[ {{{\cos }^2}x-\left( {\dfrac{1}{2}} \right) \cos x} \right] $
Adjusting the right-hand side term in ${\left( {x - y} \right)^2}$form,
$y = -2 \left[ {{{\left( {cosx - \left( {\dfrac{1}{4}} \right)} \right)}^2}\;- \left( {\dfrac{1}{{16}}} \right)} \right]$
Evaluating the right-hand side, we get
$y = \left( {\dfrac{1}{8}} \right) - 2 {\left[ {cosx - \left( {\dfrac{1}{4}} \right)} \right]^2}$
So, the maximum value is obtained when the subtraction part tends to zero in the above obtained expression.
Therefore, the maximum value of $y $ will be,
${y_{max}}\; = \dfrac{1}{8} $

Option ‘D’ is correct

Note: The given problem can also be solved by dividing the given linear first-order differential equation by $y $ thus transforming the equation into $ \dfrac{1}{y}\dfrac{{dy}}{{dx}} + 2\tan x = \dfrac{1}{y}\sin x$, Thereafter integrating the expression to obtain the required solution. We should also take care while substituting the value of x & y in the equation. One should also have the knowledge to obtain maximum from a given equation, as we obtained the ${y_{max}}$.