Suppose $y(1) = 0$ and $xy'(x)+2y(x) \geq 3x$ for all $x > 0$. How can I show them that $y(x) \geq x - 1/x^2$ for all $x\geq 1$?
I know how to find the general solution and the particular solution for $y(1)=0$ but it seems quite strange with the inequality. How would I go about doing this question that involves inequalities? Is it the same method as if the equation had an "$=$" sign?
Thank you
Hint
$$xy'(x)+2y(x) \geq 3x \,\,\, \forall x>0$$
Since $x>0$ $$x^2y'(x)+2xy(x) \geq 3x^2 $$ $$(x^2y)' \geq 3x^2 $$ $$x^2y \geq \int 3x^2=x^3+K \,\,\, \forall x>0$$ Since $x>0$ $$y \geq x+\frac K {x^2}$$ $$y(1)=0 \to 0 \geq 1+K \to K \leq -1$$ $$....$$