Question

Question:- If y(x) is a solution of [2+ sin x/1+y] dy/dx = - cos x and y(0) = 1, then find the value of y [π/2]​

Answer 1

Given that y(x) is a solution of the differential equation,

$\longrightarrow\left(\dfrac{2+\sin x}{1+y}\right)\dfrac{dy}{dx}=-\cos x$

$\longrightarrow\dfrac{2+\sin x}{1+y}\cdot\dfrac{dy}{dx}=-\cos x$

$\longrightarrow\dfrac{dy}{1+y}=-\dfrac{\cos x\ dx}{2+\sin x}$

Integrating,

$\displaystyle\longrightarrow\int\dfrac{dy}{1+y}=-\int\dfrac{\cos x\ dx}{2+\sin x}$

$\displaystyle\longrightarrow\log|1+y|=-\log|2+\sin x|+\log|C|$

Taking antilog,

$\displaystyle\longrightarrow 1+y=\dfrac{C}{2+\sin x}\quad\quad\dots(1)$

At $(x,\ y)=(0,\ 1),$

$\displaystyle\longrightarrow 1+1=\dfrac{C}{2+\sin 0}$

$\displaystyle\longrightarrow 2=\dfrac{C}{2}$

$\longrightarrow C=4$

Then (1) becomes,

$\displaystyle\longrightarrow 1+y=\dfrac{4}{2+\sin x}$

$\displaystyle\longrightarrow y=\dfrac{4}{2+\sin x}-1$

Then, at $x=\dfrac{\pi}{2},$

$\displaystyle\longrightarrow y=\dfrac{4}{2+\sin\left(\dfrac{\pi}{2}\right)}-1$

$\displaystyle\longrightarrow y=\dfrac{4}{2+1}-1$

$\displaystyle\longrightarrow y=\dfrac{4}{3}-1$

$\displaystyle\longrightarrow\underline{\underline{y=\dfrac{1}{3}}}$

Hence $\bf{y\left(\dfrac{\pi}{2}\right)=\dfrac{1}{3}}.$

Answer 2

Answer:

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