Question

Let $y = y(x)$ be the solution of the differential equation \[\cos \;x(3\sin \;x + \cos \;x + 3)dy = (1 + y\sin x(3\sin \,x + \cos \,x + 3))dx,\;0 \leqslant x \leqslant \dfrac{\pi }{2},\;y(0) = 0\]. Then, $y\left( {\dfrac{\pi }{3}} \right)$ is equal to
(A) $2{\log _e}\left[ {\dfrac{{(2\sqrt 3 + 10)}}{{11}}} \right]$
(B) $2{\log _e}\left[ {\dfrac{{(\sqrt 3 + 7)}}{2}} \right]$
(C) $2{\log _e}\left[ {\dfrac{{(3\sqrt 3 - 8)}}{4}} \right]$
(D) $2{\log _e}\left[ {\dfrac{{(2\sqrt 3 + 9)}}{6}} \right]$

Answer 1

Hint: To solve this question, we will first expand R.H.S then we will simplify it using the product rule and integrate it. Further, we will use suitable trigonometric identities to get a more simplified equation. Next, we will use differentiation, substitution, completing the square method, and rationalization to get the final answer.

Formula Used:
$(x\;dy - y\;dx = d\;xy)$
$\left[ {\sin \;x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right]$
$\left[ {\cos \;x = 1 - 2{{\sin }^2}\dfrac{x}{2}} \right]$
$\left[ {\dfrac{1}{{\cos \;x}} = \sec x} \right]$
$\left[ {\dfrac{{\sin x}}{{\cos \;x}} = \tan \;x} \right]$
$\left[ {{{\sec }^2}x = 1 + {{\tan }^2}x} \right]$
$\left[ {{{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right]$
$\left[ {\int {\dfrac{{dx}}{{{x^2} - {a^2}}} = \dfrac{1}{{2a}}\ln \left| {\dfrac{{x - a}}{{x + a}}} \right|} } \right]$
$\left[ {\ln \;a + \ln \;b = \ln \;ab} \right]$

Complete step by step Solution:
The given equation is
\[\cos \;x(3\sin \;x + \cos \;x + 3)dy = (1 + y\sin x(3\sin \,x + \cos \,x + 3))dx\]
Expanding R.H.S
\[\cos \;x(3\sin \;x + \cos \;x + 3)dy = dx + y\sin x(3\sin \,x + \cos \,x + 3)dx\]
Taking everything to L.H.S except $dx$
\[\cos \;x(3\sin \;x + \cos \;x + 3)dy - y\sin x(3\sin \,x + \cos \,x + 3)dx = dx\]
Taking \[(3\sin \,x + \cos \,x + 3)\] as common
$(\cos \,x\;dy - y\;\sin \;x\;dx)(3\sin \;x + \cos \;x + 3) = dx$
Using product rule$,$ i.e.$,$ $(x\;dy - y\;dx = d\;xy)$
$d(y - \cos \;x) = \dfrac{{dx}}{{3\sin \;x + \cos \;x + 3}}$ $\left[ {\because \cos \;x\,dy - y\sin \;x\;dx = d(y - \cos \;x)} \right]$
Integrating both sides$,$
$\int {d(y - \cos \;x)} = \int {\dfrac{{dx}}{{3\sin \;x + \cos \;x + 3}}} $
Using trigonometric identities$,$
$y\cos \;x = \int {\dfrac{{dx}}{{3\left( {2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right) + \left( {1 - 2{{\sin }^2}\dfrac{x}{2}} \right) + 3}}} $ $\left[ {\because \sin \;x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right]$$,$$\left[ {\because \cos \;x = 1 - 2{{\sin }^2}\dfrac{x}{2}} \right]$
Solving it$,$
\[y\cos \;x = \int {\dfrac{{dx}}{{6\sin \dfrac{x}{2}\cos \dfrac{x}{2} - 2{{\sin }^2}\dfrac{x}{2} + 4}}} \]
Divide numerator and denominator by ${\cos ^2}\dfrac{x}{2}$
\[y\cos \;x = \int {\dfrac{{\left( {\dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}}} \right)dx}}{{\left( {\dfrac{{6\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}} \right) - \left( {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}} \right) + \left( {\dfrac{4}{{{{\cos }^2}\dfrac{x}{2}}}} \right)}}} \]
Solving it further$,$
$y\;\cos \;x = \dfrac{{{{\sec }^2}\dfrac{x}{2}dx}}{{6\tan \dfrac{x}{2} - 2{{\tan }^2}\dfrac{x}{2} + 4{{\sec }^2}\dfrac{x}{2}}}$ $\left[ {\because \dfrac{1}{{\cos \;x}} = \sec x} \right]$$,$$\left[ {\because \dfrac{{\sin x}}{{\cos \;x}} = \tan \;x} \right]$
Using trigonometric identity,
$y\;\cos \;x = \dfrac{{{{\sec }^2}\dfrac{x}{2}dx}}{{6\tan \dfrac{x}{2} - 2{{\tan }^2}\dfrac{x}{2} + 4\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)}}$ $\left[ {\because {{\sec }^2}x = 1 + {{\tan }^2}x} \right]$
Simplifying it$,$
$y\;\cos \;x = \dfrac{{{{\sec }^2}\dfrac{x}{2}dx}}{{6\tan \dfrac{x}{2} - 2{{\tan }^2}\dfrac{x}{2} + 4 + 4{{\tan }^2}\dfrac{x}{2}}}$
$y\;\cos \;x = \dfrac{{{{\sec }^2}\dfrac{x}{2}dx}}{{6\tan \dfrac{x}{2} + 2{{\tan }^2}\dfrac{x}{2} + 4}}$
Divide numerator and denominator by 2
\[y\;\cos \;x = \dfrac{{\dfrac{1}{2}{{\sec }^2}\dfrac{x}{2}dx}}{{3\tan \dfrac{x}{2} + {{\tan }^2}\dfrac{x}{2} + 2}}\] …………..equation $(1)$
Let $\tan \dfrac{x}{2} = t$ ………………equation $(2)$
Differentiating both sides$,$
$\dfrac{1}{2}{\sec ^2}\dfrac{x}{2}dx = dt$ ………………equation $(3)$
Substituting equation $(2)$ and $(3)$ in equation $(1),$
$y\;\cos \;x = \int {\dfrac{{dt}}{{3t + {t^2} + 2}}} $
Applying completing the square method in the denominator$,$ i.e.$,$ add and subtract $\dfrac{9}{4}$ $($square of half of the coefficient of t$)$ in the denominator
$y\;\cos \;x = \int {\dfrac{{dt}}{{\left( {3t + {t^2} + \dfrac{9}{4}} \right) - \dfrac{9}{4} + 2}}} $
$y\;\cos \;x = \int {\dfrac{{dt}}{{{{\left( {t + \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}}}} $ $\left[ {\because {{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right]$
Integrating$,$
$y\;\cos \;x = \dfrac{1}{{2\left( {\dfrac{1}{2}} \right)}}\ln \left| {\dfrac{{\left( {t + \dfrac{3}{2}} \right) - \dfrac{1}{2}}}{{\left( {t + \dfrac{3}{2}} \right) + \dfrac{1}{2}}}} \right| + C$ ………………equation $(4)$ $\left[ {\because \int {\dfrac{{dx}}{{{x^2} - {a^2}}} = \dfrac{1}{{2a}}\ln \left| {\dfrac{{x - a}}{{x + a}}} \right|} } \right]$
Substituting equation $(2)$ in equation $(4)$ and solving it$,$
$y\;\cos \;x = \ln \left| {\dfrac{{\tan \dfrac{x}{2} + 1}}{{\tan \dfrac{x}{2} - 2}}} \right| + C$ ………………equation $(5)$
We are given that $y(0) = 0$
Using it in equation $(5),$
$0 = \ln \left( {\dfrac{1}{2}} \right) + C$
$C = \ln (2)$
Substituting the value of $C$ in equation $(5),$
\[y\;\cos \;x = \ln \left| {\dfrac{{\tan \dfrac{x}{2} + 1}}{{\tan \dfrac{x}{2} - 2}}} \right| + \ln (2)\]
For $x = \dfrac{\pi }{3},$
\[y\;\cos \left( {\dfrac{\pi }{3}} \right) = \ln \left| {\dfrac{{\tan \dfrac{\pi }{6} + 1}}{{\tan \dfrac{\pi }{6} - 2}}} \right| + \ln (2)\]
\[y\;\left( {\dfrac{1}{2}} \right) = \ln \left| {\dfrac{{\dfrac{1}{{\sqrt 3 }} + 1}}{{\dfrac{1}{{\sqrt 3 }} - 2}}} \right| + \ln (2)\]
On simplifying$,$
$y\left( {\dfrac{1}{2}} \right) = \ln \left( {\dfrac{{1 + \sqrt 3 }}{{1 - 2\sqrt 3 }}} \right) + \ln (2)$
Rationalizing$,$
\[y\left( {\dfrac{1}{2}} \right) = \ln \left| {\left( {\dfrac{{1 + \sqrt 3 }}{{1 - 2\sqrt 3 }}} \right)\left( {\dfrac{{1 + 2\sqrt 3 }}{{1 + 2\sqrt 3 }}} \right)} \right| + \ln (2)\]
Simplifying it further$,$
\[y\left( {\dfrac{1}{2}} \right) = \ln \left( {\dfrac{{5 + \sqrt 3 }}{{11}}} \right) + \ln (2)\]
\[y\left( {\dfrac{1}{2}} \right) = \ln \left( {\dfrac{{10 + 2\sqrt 3 }}{{11}}} \right)\] $\left[ {\because \ln \;a + \ln \;b = \ln \;ab} \right]$
Solving it to get the final answer$,$
\[y = 2\ln \left( {\dfrac{{10 + 2\sqrt 3 }}{{11}}} \right)\] or \[y = 2{\log _e}\left[ {\dfrac{{(2\sqrt 3 + 10)}}{{11}}} \right]\]

Hence, the correct option is A.

Note: The key concept to solving this type of question is to be very sure and attentive while solving it as it involves a lot of steps like integration, differentiation, rationalization, use of trigonometric identities, completing the square method, and substitution. Also, take proper care while applying the identities.