Question

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

Answer 1

Hint:\[y\left( 0 \right) = 1\] means that at the point 0 the value of the function is one and we have to find the value of the function at \[\dfrac{\pi }{2}\]. The given equation is such that we have to convert it into a partial differential equation from. Remember that integration of \[\dfrac{1}{x}dx\] is \[\log \left| x \right|\] and differentiation of \[\cos x\,\,is\,\,\sin x\] and \[\sin x\]is \[\cos x\,\] .
Ex: \[\dfrac{d}{{dx}}\sin x = \cos x\]
The log function has many different rules. One can write \[\left( {\log m + \log n} \right)\] as \[{\text{log mn and log m - log n}}\] as \[\log \dfrac{m}{n}\] similar \[{\text{log x}}\] to the house power is power \[{\text{x log x}}\].
Ex: \[\log {x^2} = 2\log x\]

Complete step by step solution:
Given: The given differential equation is
\[
  \left( {2 + \sin x} \right)\dfrac{{dy}}{{dx}} + \left( {y + 1} \right)\cos x = 0 \\
   \Rightarrow \left( {2 + \sin x} \right).dy + \left( {y + 1} \right).\cos x\,\,dx = 0 \\
 \]
Further performing calculation, we get
\[
   \Rightarrow 2 + \sin x.dy = - \left( {y + 1} \right).\cos x.dx \\
   \Rightarrow - \dfrac{1}{{\left( {y + 1} \right)}}dy = \dfrac{{\cos x}}{{2 + \sin x}}.dx \\
   \Rightarrow \dfrac{1}{{y + 1}}.dy + \dfrac{{\cos x}}{{2 + \sin x}}.dx = 0 \\
 \]
Now performing integration i.e. integrating both sides, we get
\[\int {\dfrac{1}{{y + 1}}} dy + \int {\dfrac{{\cos x}}{{2 + \sin x}} = dx = 0 ........\left( i \right)} \]
Let \[2 + \sin x = t\]
\[\therefore \]Differentiating with respect to x we get
\[
  \dfrac{d}{{dx}}\left( {2 + \sin x} \right) = \dfrac{{dt}}{{dx}} \\
   \Rightarrow \cos x\,\,dx = dt ....\left( {ii} \right) \\
 \]
Now, putting the value of \[\cos xdx = dt\] and \[2 + \sin x = t\] in the equation (i) we get
\[
  \log \left| {1 + y} \right| + {C_1}\, + \int {\dfrac{{dt}}{t}} = 0 \\
   \Rightarrow \log \left| {1 + y} \right| + {C_1} + \log \left| t \right| + {C_2} = 0 \\
 \]
Putting back the value of t.
\[ \Rightarrow \log \left| {1 + y} \right| + \log \left| {2 + \sin x} \right| = - \left( {{C_1} + {C_2}} \right)\]
Now, let \[ - \left( {{C_1} + {C_2}} \right) = \log \,C\] thus, the equation becomes.
\[\log \left| {1 + y} \right| + \log \left| {2 + \sin x} \right| = log\,C\]
Now, according to question \[y\left( 0 \right) = 1\] this means that \[y = 1\]when \[x = 0\], thus putting these values we get
\[\log \left( 2 \right) + \log \left| 2 \right| = \log C\]
\[
   \Rightarrow C = 2 + 2 \\
  \,\,\,\,\,\,\,C = 4 \\
 \] [\[\because \log M + \log N = \log z = \log \left( {MN} \right) = \log Z\]thus, \[{\text{MN = Z}}\]]
Now, the equation becomes,
\[\left( {1 + y} \right)\left( {2 + \sin x} \right) = 4\]
Now, at \[x = \dfrac{\pi }{2}\]
We get
\[\left( {1 + y} \right)\left( {2 + \sin \dfrac{\pi }{2}} \right) = 4\]
Therefore,
\[\left( {y + 1} \right)\left( {2 + 1} \right) = 4\] [Since \[\left( {\sin \dfrac{\pi }{2} = 1} \right)\]]
\[
  \left( {y + 1} \right) = \dfrac{4}{3} \\
  y = \dfrac{4}{3} - 1 \\
 \]
Therefore,
\[y = \dfrac{{4 - 3}}{3} = \dfrac{1}{3}\]
That at \[x\left( {\dfrac{\pi }{2}} \right) = \dfrac{1}{3}\] is the required answer.
Thus option (1) is correct.

Note: In this type of question students often make mistakes while performing logarithmic operation, do not make such mistakes.