Question

If \[\left( {\dfrac{{2 + \sin x}}{{1 + y}}} \right)\dfrac{{dy}}{{dx}} = - \cos x\], \[y\left( 0 \right) = 1\], then find the value of \[y\left( {\dfrac{\pi }{2}} \right)\].
A. 1
B. \[\dfrac{1}{2}\]
C. \[\dfrac{1}{3}\]
D. \[\dfrac{1}{4}\]

Answer 1

Hint: First we have to solve the given differential equation. To solve the differential equation we will separate the variables of the differential equation and integrate them. Then substitute the initial condition to find the value of the integrating constant. Then substitute \[x = \dfrac{\pi }{2}\] in the solution to calculate \[y\left( {\dfrac{\pi }{2}} \right)\].

Formula Used: Integrating formula:
\[\int {\dfrac{1}{x}dx} = \log x + c\]
Differentiate formula:
\[\dfrac{d}{{d\theta }}\left( {\sin \theta } \right) = \cos \theta \]

Complete step by step solution: Given differential equation is:
\[\left( {\dfrac{{2 + \sin x}}{{1 + y}}} \right)\dfrac{{dy}}{{dx}} = - \cos x\]
Now separate the variable of the given differential equation:
\[ \Rightarrow \dfrac{{dy}}{{1 + y}} = - \dfrac{{\cos x}}{{2 + \sin x}}dx\]
Now taking integration on both sides:
\[ \Rightarrow \int {\dfrac{{dy}}{{1 + y}}} = - \int {\dfrac{{\cos x}}{{2 + \sin x}}dx} \]
\[ \Rightarrow \int {\dfrac{{dy}}{{1 + y}}} + \int {\dfrac{{\cos x}}{{2 + \sin x}}dx} = 0\] …..(1)
Assume that, \[1 + y = z\] and \[2 + \sin x = t\]
Differentiate both sides:
\[dy = dz\] and \[\cos xdx = dt\]
Now we will substitute \[1 + y = z\], \[2 + \sin x = t\], \[dy = dz\] and \[\cos xdx = dt\] in equation (i)
\[ \Rightarrow \int {\dfrac{{dz}}{z}} + \int {\dfrac{{dt}}{t}} = 0\]
Now applying integration formula \[\int {\dfrac{1}{x}dx} = \log x + c\]
\[ \Rightarrow \log z + \log t = \log c\]
Substitute the value of z and t:
\[ \Rightarrow \log \left( {1 + y} \right) + \log \left( {2 + \sin x} \right) = \log c\]
Apply product rule of logarithm:
\[ \Rightarrow \log \left( {1 + y} \right)\left( {2 + \sin x} \right) = \log c\]
Applying anti log formula:
\[ \Rightarrow \left( {1 + y} \right)\left( {2 + \sin x} \right) = c\] ….(ii)
Substitute x = 0 and y = 1
\[ \Rightarrow \left( {1 + 1} \right)\left( {2 + \sin 0} \right) = c\]
\[ \Rightarrow 4 = c\]
Substitute c= 4 in equation (ii)
\[ \Rightarrow \left( {1 + y} \right)\left( {2 + \sin x} \right) = 4\]
Now substitute \[x = \dfrac{\pi }{2}\] in the above equation:
\[ \Rightarrow \left( {1 + y} \right)\left( {2 + \sin \dfrac{\pi }{2}} \right) = 4\]
\[ \Rightarrow \left( {1 + y} \right)\left( {2 + 1} \right) = 4\]
\[ \Rightarrow \left( {1 + y} \right)3 = 4\]
Divide both sides by 3:
\[ \Rightarrow \left( {1 + y} \right) = \dfrac{4}{3}\]
\[ \Rightarrow y = \dfrac{4}{3} - 1\]
\[ \Rightarrow y = \dfrac{1}{3}\]

Option ‘C’ is correct

Note: Student often do mistake to differentiate \[\sin x\]. They calculate that the differentiate of \[\sin x\] is \[ - \cos x\] which is incorrect. The differentiate of \[\sin x\] is \[\cos x\].