Differential Equations 2 Question 14
14. Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)$.
If $y \frac{\pi}{2}=0$, then $y \frac{\pi}{6}$ is equal to
(a) $\frac{4}{9 \sqrt{3}} \pi^{2}$
(b) $\frac{-8}{9 \sqrt{3}} \pi^{2}$
(c) $-\frac{8}{9} \pi^{2}$
(d) $-\frac{4}{9} \pi^{2}$
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Answer:
Correct Answer: 14. (c)
Solution:
- We have,
$\sin x \frac{d y}{d x}+y \cos x=4 x \Rightarrow \frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x$
This is a linear differential equation of form
$$ \frac{d y}{d x}+P y=Q $$
where $P=\cot x, Q=4 x \operatorname{cosec} x$
Now, $I F=e^{\int P d x}=e^{\int \cot x d x}=e^{\log \sin x}=\sin x$
Solution of the differential equation is
$$ \begin{array}{rlrl} & & y \cdot \sin x=\int 4 x \operatorname{cosec} x \sin x d x+C \\ \Rightarrow \quad & y \sin x=\int 4 x d x+C=2 x^{2}+C \end{array} $$
Put $x=\frac{\pi}{2}, y=0$, we get
$$ \begin{aligned} & C=-\frac{\pi^{2}}{2} \Rightarrow y \sin x=2 x^{2}-\frac{\pi^{2}}{2} \\ \text { Put } & x=\frac{\pi}{6} \\ \therefore & y \frac{1}{2}=2 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2} \\ \Rightarrow & y=\frac{\pi^{2}}{9}-\pi^{2} \Rightarrow y=-\frac{8 \pi^{2}}{9} \end{aligned} $$
Alternate Method
We have, $\sin x \frac{d y}{d x}+y \cos x=4 x$, which can be written as $\frac{d}{d x}(\sin x \cdot y)=4 x$
On integrating both sides, we get
$$ \begin{aligned} & \int \frac{d}{d x}(\sin x \cdot y) \cdot d x=\int 4 x \cdot d x \\ \Rightarrow & y \cdot \sin x=\frac{4 x^{2}}{2}+C \Rightarrow y \cdot \sin x=2 x^{2}+C \end{aligned} $$
Now, as $y=0$ when $x=\frac{\pi}{2}$
$$ \begin{aligned} & \therefore & C & =-\frac{\pi^{2}}{2} \\ & \Rightarrow & y \cdot \sin x & =2 x^{2}-\frac{\pi^{2}}{2} \end{aligned} $$
Now, putting $x=\frac{\pi}{6}$, we get
$$ y \frac{1}{2}=2 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2} \Rightarrow y=\frac{\pi^{2}}{9}-\pi^{2}=-\frac{8 \pi^{2}}{9} $$
