Question
Let $$y - y\left( x \right)$$ be the solution of the differential equation $$\sin \,x\frac{{dy}}{{dx}} + y\,\cos \,x = 4x,\,x \in \left( {0,\,\,\pi } \right).$$ If $$y\left( {\frac{\pi }{2}} \right) = 0,$$ then $$y\left( {\frac{\pi }{6}} \right)$$ is equal to :
A.
$$\frac{{ - 8}}{{9\sqrt 3 }}{\pi ^2}$$
B.
$$ - \frac{8}{9}{\pi ^2}$$
C.
$$ - \frac{4}{9}{\pi ^2}$$
D.
$$\frac{4}{{9\sqrt 3 }}{\pi ^2}$$
Answer :
$$ - \frac{8}{9}{\pi ^2}$$
Solution :
Consider the given differential equation the
$$\eqalign{
& \sin \,xdy + y\,\cos \,xdx = 4xdx \cr
& \Rightarrow d\left( {y.\sin \,x} \right) = 4xdx \cr} $$
Integrate both sides
$$\eqalign{
& \Rightarrow y.\sin \,x = 2{x^2} + C\,.....(1) \cr
& \Rightarrow y\left( x \right) = \frac{{2{x^2}}}{{\sin \,x}} + c\,.....(2) \cr} $$
$$\because $$ equation (2) passes through $$\left( {\frac{\pi }{2},\,\,0} \right)$$
$$ \Rightarrow 0 = \frac{{{\pi ^2}}}{2} + C\,\,\, \Rightarrow C = - \frac{{{\pi ^2}}}{2}$$
Now, put the value of $$C$$ in (1)
Then, $$y\,\sin \,x = 2{x^2} - \frac{{{\pi ^2}}}{2}$$ is the solution
$$\therefore y\left( {\frac{\pi }{6}} \right) = \left( {2.\frac{{{\pi ^2}}}{{36}} - \frac{{{\pi ^2}}}{2}} \right)2 = - \frac{{8{\pi ^2}}}{9}$$