Question

If \[\;y = y(x)\] is the solution of the differential equation \[\,\left( {\dfrac{{2 + \sin x}}{{y + 1}}} \right)\dfrac{{dy}}{{dx}} + \cos x = 0\].
With \[y(0) = 1\] then \[y(\dfrac{\pi} {2} )\;\_\_\].

Answer 1

Hint: Separate dy and dx then integrate the equation. We use the variable separable method.
Given :-
\[\left( {\dfrac{{2 + \sin x}}{{y + 1}}} \right)\dfrac{{dy}}{{dx}} + \cos x = 0\]
After transposing we get,
\[\left( {\dfrac{1}{{y + 1}}} \right)dy = - \left( {\dfrac{{\cos x}}{{2 + \sin x}}} \right)dx\]
On integrating we get,
\[\int {\left( {\dfrac{1}{{y + 1}}} \right)dy = - \int {\left( {\dfrac{{\cos x}}{{2 + \sin x}}} \right)dx} } \,\,\,...({\text{i}})\]
First we will solve this,
\[\int {\left( {\dfrac{{\cos x}}{{2 + \sin x}}} \right)dx} \]
Let \[\sin x = t\]
Then \[\cos xdx = dt\]
From the above two equations the integral becomes,
\[\int {\left( {\dfrac{1}{{2 + t}}} \right)dt = \ln (2 + t)} \,\, + \ln {c_1}\]
On putting the value of \[{\text{ }}t\],
\[\log (2 + t) = \log (2 + \sin x)\]
On putting these values in equation (i), we get equation (i) as ,
\[\ln (y + 1) = - \ln (2 + \sin x) + \ln c\]
Then we got an equation as,
\[y = \dfrac{c}{{2 + \sin x}}\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(\ln a - \ln b = \ln \dfrac{a}{b})\]
From the question,
\[y(0) = 1\]
So,
\[
  {\text{1 = }}\dfrac{c}{{2 + 0}} - 1\,\,\,\,\,\,\,\,\,\,\,......(\sin 0 = 0) \\
  c = 4 \\
\]
The equation is ,
\[y = \dfrac{4}{{2 + \sin x}}\, - 1\,\,\,\]
We have asked to find \[y\left( {\dfrac{\pi }{2}} \right)\],
So,
\[y\left( {\dfrac{\pi }{2}} \right) = \,\dfrac{4}{{2 + 1}}\, - 1 = \dfrac{1}{3}\,\,\,\,\,\,\,\,\,......(\sin \left( {\dfrac{\pi }{2}} \right) = 1)\]
Hence the correct option is A.

Note: In these types of questions first open the integral by variable separation method then when the equation is obtained, get the value of constant from the information provided in the question then after getting completely , get the answer which is asked in question by putting the value of constant.