пожалуйста помогите решить)

[tex]x\cdot f'(x)=f(x)+4, \ \ f(x)=x- \frac{1}{x} \\\\ f'(x)=(x- \frac{1}{x} )'=(x)'-( \frac{1}{x} )'=1- \frac{1'\cdot x-1\cdot x'}{x^2}=1- (-\frac{1}{x^2})=\\\\ =1+ \frac{1}{x^2}= \frac{x^2+1}{x^2}\\\\\\ x\cdot( \frac{x^2+1}{x^2})= \frac{x^2-1}{x}+4\\\\ \frac{x^2+1}{x}= \frac{x^2-1+4x}{x}\\\\ \frac{x^2+1-x^2+1-4x}{x}=0\\\\ \frac{-4x+2}{x}=0 \\\\ x\neq0\\\\ -4x+2=0\\\\ -4x=-2\\\\ x= \frac{1}{2} [/tex]