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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.

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

Answers (1)

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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