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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 10

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 10

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Answer: \left(e^{y}+1\right) \sin x=c

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

Given: \left(e^{y}+1\right) \cos x d x+e^{y} \sin x d y=0

Solution: \left(e^{y}+1\right) \cos x d x+e^{y} \sin x d y=0

        e^{y} \sin x d y=-\left(e^{y}+1\right) \cos x d x

        \frac{-e^{y}}{\left(e^{y}+1\right)} d y=\frac{\cos x}{\sin x} d x

        Integrating both sides

        \begin{aligned} &\int \frac{-e^{y}}{\left(e^{y}+1\right)} d y=\int \frac{\cos x}{\sin x} d x \\\\ &\text { Put } e^{y}+1=t \Rightarrow e^{y} d y=d t \\\\ &\sin x=u \Rightarrow d x \cos x=d u \end{aligned}

        \begin{aligned} &-\int \frac{d t}{t}=\int \frac{d u}{u} \\\\ &-\log |t|=\log |u|+\log c \\\\ &-\log \left|e^{y}+1\right|=\log |\sin x|+\log c \end{aligned}

        \begin{aligned} &-\log \left|e^{y}+1\right|-\log |\sin x|=\log c \\\\ &-\left[\log \left(e^{y}+1\right)(\sin x)\right]=\log c \\\\ &{\left[\log \left(e^{y}+1\right)(\sin x)\right]=-\log c} \\\\ &\left(e^{y}+1\right) \sin x=c \end{aligned}

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