Prove: $f(t) leq K_1 e^{K_2(t-a)}+frac{varepsilon}{K_2} e^{K_2(t-a)-1}$
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Let $f$ be a non-negative function that satisfies $$f(t) leq K_1 + varepsilon(t-a)+K_2 int_a^t f(s)ds$$ for $aleq t leq b, quad K_1,K_2,varepsilon in mathbb{R}$ . Prove: $$f(t) leq K_1 e^{K_2(t-a)}+frac{varepsilon}{K_2} e^{K_2(t-a)-1}$$ My attempt: Let $$U(t)=K_1 + varepsilon(t-a)+K_2 int_a^t f(s)ds, quad U'(t)=varepsilon t +K_2f(t)$$ then $$U'(t) leq varepsilon t + K_2 U(t) iff U'(t)-K_2U(t) leq varepsilon t quad Bigg| cdot e^{-K_2(t-a)}$$ $$Big[e^{-K_2(t-a)}U(t)Big]' leq int varepsilon t cdot e^{-K_2(t-a)} dt +c iff U(t) leq -frac{varepsilon}{K_2^2}(K_2t+1)+ccdot e^{K_2(t-a)} iff$$ $$int_a^t f(s) ds leq K_1-varepsilon (t-a)-frac{varepsilon}{K_2^2}(K_2t+1)+ccdot e^{K_2(t-a)}$$ but I can't derive the inequality that has to be proven. Is my $U(t)$ a b
