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  • If y(x) is a solution of \frac{2 + sin x}{1 +y}frac{dy}{dx}=-cos x and y(0) = 1, then find the value of y=frac{\pi}{2}

If y(x) is a solution of \frac{2 + \sin x}{1 +y}\frac{dy}{dx}=-\cos x and y(0) = 1, then find the value of y=\frac{\pi}{2}

Answers (1)

Given:

\frac{2+\sin x}{1+y} \frac{d y}{d x}=-\cos x

To find: Solution of the differential equation

 

Rewriting the given equation as,

\frac{d y}{1+y}=\frac{-\cos x d x}{2+\sin x}

Integrating on both sides,

\\ \int \frac{d y}{1+y}=\int \frac{-\cos x d x}{2+\sin x} \\ \ln (1+y)=\int \frac{-\cos x d x}{2+\sin x}

Let sinx=t and cos xdx= dt

\begin{array}{l} \text { formula: } \frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=\cos \mathrm{x} \\ \ln (1+\mathrm{y})=\int \frac{-\mathrm{dt}}{2+\mathrm{t}} \\ \text { Formula: } \int \frac{\mathrm{dt}}{\mathrm{t}}=\ln \mathrm{t} \end{array}

ln(1+y)=-ln(2+t)+logc

ln(1+y)+ln(2++sinx)=logc

(1+y)(2+sinx)=c

When x=0 and y=1

\begin{array}{l} 1=\frac{c}{2+\sin 0}-1 \\ 1+1=\frac{c}{2} \end{array}

c=4

\begin{array}{l} \Rightarrow \mathrm{y}=\frac{4}{2+\sin \mathrm{x}}-1 \\ \text { If } \mathrm{x}=\pi / 2 \text { then, } \\ \mathrm{y}=\frac{4}{2+\sin \frac{\pi}{2}}-1 \end{array}

\begin{array}{l} y=\frac{4}{2+1}-1 \\\\ y=\frac{4}{3}-1 \\\\ y=\frac{1}{3} \end{array}


 

Posted by

Ravindra Pindel

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