Need a step-by-step solution to this

Question

Given a first order differential equation $$\frac{1}{x}\frac{dy}{dx} - \frac{2}{x^2}y = x\cos x,$$ obtain the general solution using the method of integrating factor

Answer 1

HINTS:

Write $$\frac{1}{x}\frac{dy}{dx} - \frac{2}{x^2}y = x\cos x\implies \frac{dy}{dx} - \frac{2}{x}y = x^2\cos x$$ Now let $$P(x)=-\frac2x,\quad Q(x)=x^2\cos x$$ and your integrating factor is $$I(x)=\exp\left(\int P(x)\,dx\right)$$ Can you continue?

Answer 2

$$\frac{1}{x}\frac{dy}{dx} - \frac{2}{x^2}y = x\cos x,$$ $$\mu\frac{1}{x}\frac{dy}{dx} - \mu\frac{2}{x^2}y = \mu x\cos x,$$

$$(\mu\frac{1}{x})' = -\mu\frac{2}{x^2}$$ Integrate to get $\mu$ $$z' = -z\frac{2}{x}$$ $$\ln z= -2 \ln x$$ $$\mu\frac{1}{x}= x^{-2 }$$ $$\mu= x^{-1 }$$

Multiply both side by $\mu$

$$\mu\frac{1}{x}\frac{dy}{dx} - \mu\frac{2}{x^2}y = \mu x\cos x,$$ $$\frac{1}{x^2}y' - \frac{2}{x^3}y = \cos x,$$ $$(\frac{1}{x^2}y )' = \cos x$$ Integrate $$(\frac{1}{x^2}y ) = \sin x+K$$ $$y = x^2(\sin x+K)$$