Ecuaciones Diferenciales por método de Variación de Parametros

Dada la siguiente ecuación: $$y\,'+ \frac{1}{x}y=3\cos 2x$$ $$y=u(x) \cdot v(x)$$ Procedemos a calcular el factor integrante $$u(x)=e^{-\displaystyle\int p(x)}$$ $$u(x)=e^{-\displaystyle\int\tfrac{1}{x}}$$ $$u(x)=\frac{1}{x}$$ $$v(x)=\displaystyle\int \tfrac{g(x)}{u(x)}$$ $$v(x)=\displaystyle\int \tfrac{3\cos 2x}{\tfrac{1}{x}}$$ $$v(x)= \displaystyle\int x3\cos 2x dx $$ $$3\displaystyle\int x \cos 2x$$ $$t=x \,\,\,\,\,\, dt=dx$$ $$ds=\cos 2x \,\,\,\,\,\, s=\frac{1}{2} \sin 2x $$ $$3\left( \frac{x}{2} \sin 2x - \displaystyle\int \tfrac{1}{2} \sin 2x \, dx\right)$$ $$3\left( \frac{x}{2} \sin 2x - +\tfrac{1}{4} \cos 2x + c\right)$$ $$v(x)=\frac{3x}{2}\sin 2x+\frac{3}{4} \cos 2x +c$$ $$y=u(x)\cdot v(x)$$ $$y=\frac{1}{x}\left( \frac{3x}{2}\sin 2x+\frac{3}{4} \cos 2x +c\right)$$ $$y=\frac{3}{2}\sin 2x+\frac{1}{4x}+\frac{c}{x}$$

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