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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 41 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 41 Maths Textbook Solution.

Answers (1)

Answer:y\cos x=\frac{\cos 2x}{2}+c

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

Given:\frac{dy}{dx}-y\tan x=-2\sin x

Solution:\frac{dy}{dx}-y\tan x=-2\sin x

\frac{dy}{dx}+\left ( -\tan x \right )y=-2\sin x

The above equation look like,

\frac{dy}{dx}+py=q

Where p = -tan x and q = -2 sin x

Integrating factor = e^{\int px}

=e^{\int -\tan x\: dx}

=e-^{\int \tan x\: dx}

we have \int \tan x\: dx=log\left ( \sec x \right )+c

\begin{aligned} &\text { I. } F=e^{-\log (s e c x)} \\ &\begin{array}{l} \text { I.F }=e^{-\log \left(\frac{1}{\operatorname{secx}}\right)} \\ \text { I.F }=\frac{1}{\sec x} \quad\left[\because \mathrm{e}^{\log x}=\mathrm{x}\right] \\ \therefore \text { I. } F=\operatorname{Cos} \mathrm{x} \end{array} \end{aligned}

Hence, now the solution of differential equation is,

\begin{aligned} &\mathrm{y}(\mathrm{I} . \mathrm{F})=\int(\mathrm{q} \times \mathrm{I} \cdot \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\ &\mathrm{y}(\operatorname{Cos} \mathrm{x})=\int(-2 \sin \mathrm{x} \cdot \cos \mathrm{x}) \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=-\int \sin 2 x \mathrm{dx}+\mathrm{C} \\ &\mathrm{y} \cos \mathrm{x}=\frac{\cos 2 x}{2}+c \end{aligned}

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