Question

Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]

Answer 1

We have, 
\[x\frac{dy}{dx} + y = x \cos x + \sin x\]
\[ \Rightarrow \frac{dy}{dx} + \frac{1}{x}y = \cos x + \frac{\sin x}{x} . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = \tan x\text{ and }Q = x^2 \cot x + 2x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\frac{1}{x}dx} \]
\[ = e^{log x} = x\]
\[\text{ Multiplying both sides of (1) by I . F . }= x,\text{ we get }\]
\[x\left( \frac{dy}{dx} + \frac{1}{x}y \right) = x\left( \cos x + \frac{\sin x}{x} \right)\]
\[ \Rightarrow x\left( \frac{dy}{dx} + \frac{1}{x}y \right) = x \cos x + \sin x\]
Integrating both sides with respect to x, we get
\[xy = \int x \cos x dx + \int\sin x dx + C\]
\[ \Rightarrow xy = \left[ x \sin x - \int1\left( \sin x \right)dx \right] - \cos x + C\]
\[ \Rightarrow xy = x \sin x + \cos x - \cos x + C\]
\[ \Rightarrow xy = x \sin x + C . . . . . \left( 2 \right)\]
Now, 
\[y\left( \frac{\pi}{2} \right) = 1\]
\[ \therefore 1 \times \frac{\pi}{2} = \frac{\pi}{2}\sin\frac{\pi}{2} + C\]
\[ \Rightarrow C = 0\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[xy = x \sin x\]
\[ \Rightarrow y = \sin x\]
\[\text{ Hence, }y = \sin x\text{ is the required solution .}\]