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I would like to solve $4u(t)+\int_0^t\sin(t-s)u(s)ds=5t, \ t\geqslant 0$.

Any ideas on how to approach this equation?

J. M. ain't a mathematician
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user339135
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  • Maybe differentiate with respect to $t$ and try to solve for $u(t)$? You might have to use leibniz rule for the integral. – MrYouMath May 15 '16 at 08:58
  • How many times is $u$ differentiable? If twice, then go with answers below –  May 15 '16 at 09:09

2 Answers2

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Hint. Assume $u(\cdot)$ is continuous over $[0,\infty)$. Then one may differentiate the initial equation twice, using the Leibniz integral rule getting

$$ u''(t)+u(t)=\frac{5}4t, \quad t\geqslant 0, $$

which can be classically solved using $u(0)=0$ and $u'(0)=\dfrac54$.

Olivier Oloa
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  • The first derivative is, I think: $;4u'+\int_0^t\cos(t-s)u(s)ds=5;$ , and the second one is:$$4u''-\overbrace{\int_0^t\sin(t-s)u(s)ds}^{-4u+5t}=0\implies 4u''+4u=-5t$$Besides the sign, how did you get $;5u;$ there, please? – DonAntonio May 15 '16 at 09:16
  • @Joanpemo Typo corrected. Thank you very much! – Olivier Oloa May 15 '16 at 09:22
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    My pleasure...and I also have a typo: it is, I think, $;5t;$ on the right side, not $;-5t;$ . – DonAntonio May 15 '16 at 09:25
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Through Laplace transform the integral equation becomes $$ 4U(s)+\frac{1}{s^2+1}U(s)=\frac{5}{s^2} $$ that is $$ U(s)=\frac{5}{s^2}\frac{s^2+1}{4s^2+5}=\frac{1}{s^2}+\frac{1}{4s^2+5} $$ and then $$ u(t)=t+\frac{1}{2\sqrt{5}}\sin\left(\frac{\sqrt{5}}{2}\;t\right)\qquad t\ge 0 $$

alexjo
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