Question

Let \[y = y\left( x \right)\] be the solution of the differential equation \[\sin x\dfrac{{dy}}{{dx}} + y\cos x = 4x,\;x \in (0,\pi ).\] If \[y\left( {\dfrac{\pi }{2}} \right) = 0\] , then \[y\left( {\dfrac{\pi }{6}} \right)\] is equal to
A. $ - \dfrac{8}{9}{\pi ^2} $
B. $ - \dfrac{4}{9}{\pi ^2} $
C. $ \dfrac{4}{{9\sqrt 3 }}{\pi ^2} $
D. $ - \dfrac{8}{{9\sqrt 3 }}{\pi ^2} $

Answer 1

Hint: First of all find the solution of the given differential equation which is of first order. And we know that to solve the first order differential equation, we have to find its integrating factor and then put it into the general solution form. After finding the solution, find the value of constant by putting the given value in the solution and after getting the final solution, find the required value.

Complete step-by-step answer:
The given differential equation \[\sin x\dfrac{{dy}}{{dx}} + y\cos x = 4x,\;x \in (0,\pi )\] is a first order differential equation we can express the differential equation as
 \[ \Rightarrow \sin x\dfrac{{dy}}{{dx}} + y\cos x = 4x\]
Dividing both sides with $ \sin x $ , we will get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} + y\cot x = \dfrac{{4x}}{{\sin x}}\]
Here we get the standard form of first order differential equation, that is $ y' + p(x)y = q(x) $
Now, find its solution, we have to find its integration factor first as follows
 $
  I.F. = {e^{\int {p(x)dx} }} \\
   = {e^{\int {\cot xdx} }} \\
   = {e^{\ln \sin x}} \\
   = \sin x \\
  $
Now, solution of first order differential equation is given as
 $ y \times I.F. = \int {I.F. \times q(x)dx} $
Putting required values, we will get
 $
   \Rightarrow y\sin x = \int {\sin x \times \dfrac{{4x}}{{\sin x}}dx} \\
   \Rightarrow y\sin x = \int {4xdx} \\
   \Rightarrow y\sin x = \dfrac{{4{x^2}}}{2} + c \\
   \Rightarrow y\sin x = 2{x^2} + c \;
  $
Therefore solution of the differential equation is $ y\sin x = 2{x^2} + c $ but we have to find the value of “c”, so we will put the given value
 \[
   \Rightarrow y\left( {\dfrac{\pi }{2}} \right) = 0 \\
   \Rightarrow 0 \times \sin \left( {\dfrac{\pi }{2}} \right) = 2{\left( {\dfrac{\pi }{2}} \right)^2} + c \\
   \Rightarrow c = - \dfrac{{{\pi ^2}}}{2} \;
 \]
Therefore solution is given as
 $ y\sin x = 2{x^2} - \dfrac{{{\pi ^2}}}{2} $
We have to find \[y\left( {\dfrac{\pi }{6}} \right)\]
 \[
   \Rightarrow y\sin \left( {\dfrac{\pi }{6}} \right) = 2{\left( {\dfrac{\pi }{6}} \right)^2} - \dfrac{{{\pi ^2}}}{2} \\
   \Rightarrow y \times \dfrac{1}{2} = \dfrac{{{\pi ^2}}}{{18}} - \dfrac{{{\pi ^2}}}{2} \\
   \Rightarrow \dfrac{y}{2} = \dfrac{{{\pi ^2}}}{2}\left( {\dfrac{1}{9} - 1} \right) \\
   \Rightarrow y = {\pi ^2}\left( { - \dfrac{8}{9}} \right) \\
   \Rightarrow y = - \dfrac{8}{9}{\pi ^2} \;
 \]
Therefore \[ - \dfrac{8}{9}{\pi ^2}\] is the required value.
So, the correct answer is “Option A”.

Note: When finding the integrating factor make sure that in integrating factor you have not added the constant part after integration. Also when you put the value of the constant in the solution of the differential equation it becomes a particular solution for that differential equation. And put the value of the variables carefully to avoid errors in the solution.