Case Study 1: Read the following passage and answer the questions given below.
Consider the differential equation

 

 

 

Answer the following questions which are based on above information. 

(i) Find the values of P and Q, if the given differential equation can be written in the form of $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}.$

(ii) Find the integrating factor of the differential equation. 

(iii) Find the solution of the differential equation.

Case Study 1: Answer

(i) Given, differential equation is 

$\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y} \tan \mathrm{x}=\sin \mathrm{x}$

Which is a linear differential equation of the form 

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}. \\ \mathrm{Here}, \mathrm{P}=2\tan \mathrm{x} \mathrm{and} \mathrm{Q}=\sin \mathrm{x} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{IF}={\mathrm{e}}^{\int \mathrm{Pdx}}={\mathrm{e}}^{\int 2\tan \mathrm{x} \mathrm{dx}} \\ ={\mathrm{e}}^{2 \int \tan \mathrm{x} \mathrm{dx}}={\mathrm{e}}^{ 2\log \left|\sec \mathrm{x}\right|} \\ ={\mathrm{e}}^{\log {\sec }^2\mathrm{x}}={\sec }^2\mathrm{x} [\because {\mathrm{e}}^{\log \mathrm{f}\left(\mathrm{x}\right)}=\mathrm{f}(\mathrm{x}\left)\right] \end{gathered}$$(iii)  The solution of differential equation is given by

$$\begin{gathered} \left(\mathrm{iii}\right) \mathrm{y}.{\sec }^2\mathrm{x}= \int {\sec }^2\mathrm{x}.\sin \mathrm{x} \mathrm{dx}+\mathrm{C} \\ \Rightarrow \mathrm{y}.{\sec }^2\mathrm{x}=\int \frac{\sin \mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{dx}+\mathrm{C} \\ \mathrm{Put} \cos \mathrm{x}=\mathrm{t}, \mathrm{then}-\sin \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ \therefore \mathrm{y}.{\sec }^2\mathrm{x}=-\int \frac{\mathrm{dt}}{{\mathrm{t}}^2} +c \\ \Rightarrow \mathrm{y}.{\sec }^2\mathrm{x}=\frac{1}{\mathrm{t}}+c \\ \Rightarrow \mathrm{y}.{\sec }^2\mathrm{x}=\frac{1}{\cos \mathrm{x}}+c [\because \mathrm{t}=\cos \mathrm{x}] \\ \Rightarrow \mathrm{y}.{\sec }^2\mathrm{x}=\sec \mathrm{x}+c \\ \Rightarrow \mathrm{y}=\frac{1}{\sec \mathrm{x}}+\frac{c}{{\sec }^2\mathrm{x}} \\ [\mathrm{dividing} \mathrm{each} \mathrm{term} \mathrm{by} {\sec }^2\mathrm{x}] \\ \Rightarrow \mathrm{y}=\cos \mathrm{x}+c.{\cos }^2\mathrm{x} \end{gathered}$$