Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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 \[y(\dfrac{\pi} {2} )\;\_\_\].

seo-qna
Last updated date: 24th Jul 2024
Total views: 453k
Views today: 4.53k
Answer
VerifiedVerified
453k+ views
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.