#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 43  Maths Textbook Solution.

Answer: $y\: \cos x=\frac{e^{x}}{2}\left ( \sin x+\cos x \right )+C$

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

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

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

Compare with $\frac{dy}{dx}+p\: y=q$  ,  we get,

\begin{aligned} &P=-\tan x \text { and } q=e^{x} \\ &\text { Now, } I . F=e^{\int-\tan x d x} \\ &=e^{-\log |(s e c x)|} \\ &=e^{\log \left(\frac{1}{\sec }\right)} \quad\left[\because e^{\log x}=x\right] \\ &\text { I. } F=\frac{1}{\sec x} \\ &=\operatorname{Cos} x \end{aligned}

Now the solution is,

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