SATHEE: Differential Equations 2 Question 14

Differential Equations 2 Question 14

14. Let $y = y (x)$ be the solution of the differential equation $\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π)$ .

If $y \frac{π}{2} = 0$ , then $y \frac{π}{6}$ is equal to

(a) $\frac{4}{9 \sqrt{3}} π^{2}$

(b) $\frac{- 8}{9 \sqrt{3}} π^{2}$

(d) $- \frac{4}{9} π^{2}$

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Answer:

Correct Answer: 14. (c)

Solution:

We have,

$\sin x \frac{d y}{d x} + y \cos x = 4 x \Rightarrow \frac{d y}{d x} + y \cot x = 4 x cosec x$

This is a linear differential equation of form

$\frac{d y}{d x} + P y = Q$

where $P = \cot x, Q = 4 x cosec x$

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

$\begin{aligned} y \cdot \sin x = \int 4 x cosec x \sin x d x + C \\ \Rightarrow & y \sin x = \int 4 x d x + C = 2 x^{2} + C \end{aligned}$

Put $x = \frac{π}{2}, y = 0$ , we get

$\begin{aligned} C = - \frac{π^{2}}{2} \Rightarrow y \sin x = 2 x^{2} - \frac{π^{2}}{2} \\ Put & x = \frac{π}{6} \\ ∴ & y \frac{1}{2} = 2 \frac{π^{2}}{36} - \frac{π^{2}}{2} \\ \Rightarrow & y = \frac{π^{2}}{9} - π^{2} \Rightarrow y = - \frac{8 π^{2}}{9} \end{aligned}$

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 $x = \frac{π}{2}$

$\begin{aligned} ∴ & C & = - \frac{π^{2}}{2} \\ \Rightarrow & y \cdot \sin x & = 2 x^{2} - \frac{π^{2}}{2} \end{aligned}$

Now, putting $x = \frac{π}{6}$ , we get

$y \frac{1}{2} = 2 \frac{π^{2}}{36} - \frac{π^{2}}{2} \Rightarrow y = \frac{π^{2}}{9} - π^{2} = - \frac{8 π^{2}}{9}$

Differential Equations 2 Question 14

14. Let $y = y (x)$ be the solution of the differential equation $\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π)$ .

Answer:

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Differential Equations 2 Question 14

14. Let y=y(x) be the solution of the differential equation sin⁡xdydx+ycos⁡x=4x,x∈(0,π).

Answer:

Alternate Method

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14. Let $y = y (x)$ be the solution of the differential equation $\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π)$ .