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Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 42  Maths Textbook Solution.

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Answer: y\: \cos \, x=e^{x}+C

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

Given:\frac{dy}{dx}-y\tan x=e^{x}\sec x

Solution:\frac{dy}{dx}-y\tan x=e^{x}\sec x

Comparing with,

\begin{aligned} &\frac{\mathrm{dy}}{\text { dx }}+\mathrm{p} \mathrm{y}=\mathrm{q} \ldots \text { (I) we get, }\\ &P=-\tan x \text { and } q=e^{x} \sec x\\ &\text { Now, I.F }=\mathrm{e}^{\int-\tan \mathrm{x} \mathrm{d} \mathrm{x}}\\ &=\mathrm{e}^{-\log |(\operatorname{secx})|}\\ &=\mathrm{e}^{\log \left(\frac{1}{\operatorname{secx}}\right)}\\ &=\mathrm{e}^{\log (\cos x)} \quad\left[\because \mathrm{e}^{\log x}=\mathrm{x}\right]\\ &=\operatorname{Cos} \mathrm{x} \end{aligned}

So, the solution is,

\begin{aligned} &\mathrm{y} \mathrm{I} . \mathrm{F}=\int(\mathrm{q} \mathrm{x} \mathrm{I} \cdot \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\ &\mathrm{y} \operatorname{Cos} \mathrm{x}=\int\left(\cos \mathrm{x} \cdot \mathrm{e}^{\mathrm{x}} \sec \mathrm{x}\right) \mathrm{dx}+\mathrm{C} \\ &=\int \cos \mathrm{x} \cdot \mathrm{e}^{x} \cdot \frac{1}{\cos x} \mathrm{dx}+\mathrm{C} \\ &=\int \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\mathrm{c} \quad\left[\because \int \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{\mathrm{x}}+\mathrm{C}\right] \\ &\therefore \mathrm{y} \operatorname{Cos} \mathrm{x}=\mathrm{e}^{\mathrm{x}}+\mathrm{C} \end{aligned}

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