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 bad choice?
integration differential-equations recurrence-relations
asked Nov 17 at 12:47
Jevaut
54310
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up vote
1
down vote
favorite
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 bad choice?
integration differential-equations recurrence-relations
asked Nov 17 at 12:47
Jevaut
54310
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 bad choice?
integration differential-equations recurrence-relations
asked Nov 17 at 12:47
Jevaut
54310
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 bad choice?
integration differential-equations recurrence-relations
integration differential-equations recurrence-relations
asked Nov 17 at 12:47
Jevaut
54310
asked Nov 17 at 12:47
Jevaut
54310
asked Nov 17 at 12:47
Jevaut
54310
asked Nov 17 at 12:47
Jevaut
54310
asked Nov 17 at 12:47
Jevaut
54310
54310
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