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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 58

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 58

Answers (1)

Answer: \frac{1}{3}

Hint: Separate the terms of x and y and then integrate them.

Given: IF y(x) is a solution of the differential equation \left(\frac{2+\sin x}{1+y}\right) \frac{d y}{d x}=-\cos x \text { and } y(0)=1 then find y\left ( \frac{\pi }{2} \right )

Solution:

        \begin{aligned} &\left(\frac{2+\sin x}{1+y}\right) \frac{d y}{d x}=-\cos x \\\\ &\Rightarrow \frac{d y}{1+y}=\frac{-\cos x}{2+\sin x} d x \end{aligned}

          Integrating both sides

          \begin{aligned} &\Rightarrow \int \frac{d y}{1+y}=-\int \frac{\cos x}{2+\sin x} d x \\\\ &\Rightarrow \log |1+y|=-\log |2+\sin x|+\log c \\\\ &{\left[\begin{array}{l} \text { Put } \\ \sin x=t \\ \cos x d x=d t \\ \int \frac{1}{t} d t=\log |t|+c \end{array}\right]} \end{aligned}

        \begin{aligned} &\Rightarrow \log |y+1|+\log |2+\sin x|=\log c \\\\ &\Rightarrow \log |(y+1)(2+\sin x)|=\log c \\\\ &\Rightarrow(y+1)(2+\sin x)=c \end{aligned}            ..........(1)

        According to given: y(0) = 1\; at\; x=0, y = 1

        \text { (2) }(2+\sin 0)=c \Rightarrow c=4

        Put in (1)

        \Rightarrow(y+1)(2+\sin x)=4                                .............(2)

        Now we have to find y\left ( \frac{\pi }{2} \right )i.e. value of y at x= \frac{\pi }{2}

        Put in (2)

        \begin{aligned} &(y+1)\left(2+\sin \frac{\pi}{2}\right)=4 \\\\ &\Rightarrow(y+1)(2+1)=4 \\\\ &{\left[\therefore \sin \frac{\pi}{2}=1\right]} \end{aligned}

        \begin{aligned} &\Rightarrow 3(y+1)=4 \\\\ &\Rightarrow 3 y+3=4 \Rightarrow 3 y=1 \Rightarrow y=\frac{1}{3} \\\\ &y\left(\frac{\pi}{2}\right)=\frac{1}{3} \end{aligned}

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