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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 40 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 40 Maths Textbook Solution.

Answers (1)

Answer: y\, \cos ec\: x+\cot \: x=c

Hint: you must know the rules of solving differential equation and integrations.

Given: \frac{dy}{dx}-y\cot x=\cos ec\: x

Solution:\frac{dy}{dx}-y\cot x=\cos ec\: x

The above equation is in form of

\frac{dy}{dx}+p\: y=q

Where p = -cot x and q = cosec x

Integrating factor = e^{\int px}

Considering \int p\: d\: x

                \begin{aligned} \Rightarrow & \int p d x=-\int \cot x d x \\ &=-\log |\sin x| &\left[\because e^{\log x}=x\right] \\ & \therefore \mathrm{e}^{\int p d x}=e^{-\log |\sin x|}=e^{\log (\sin x)^{-1}} \\ &=\sin x^{-1}=\frac{1}{\sin x}=\operatorname{cosec} x \end{aligned}

\therefore \text { Integrating factor I.F }=\operatorname{cosec} \mathrm{x}

Now, General solution is,

\begin{aligned} &\mathrm{y}(\mathrm{I} . \mathrm{F})=\int \mathrm{q}(\mathrm{I} . \mathrm{F}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=\int \operatorname{cosec} \mathrm{x} \operatorname{cosec} \mathrm{x} \mathrm{dx}+\mathrm{c} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=\int \operatorname{cosec} \mathrm{x}^{2} \mathrm{dx}+\mathrm{c} \\ &\mathrm{y} \operatorname{cosec} \mathrm{x}=-\cot \mathrm{x}+\mathrm{C} \\ &\Rightarrow \mathrm{y} \operatorname{cosec} x+\cot \mathrm{x}=\mathrm{c} \end{aligned}

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