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

$y\left( 0 \right) = 1$ 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.