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Prove using differentiation that $\phi(t)= \int^{t}_{t_0} X(t,\lambda) f(\lambda) \,d \lambda $, is the solution of the system below, where $X(t,\lambda) =X(t)X^{-1}(\lambda)$ and $X$ is fundamental matrix of this system:

$$ \begin{cases} x'=A(t)x + f(t)\\ x(t_0)=0 \end{cases}.$$

So I have that $$x'= \frac{d}{dt}\phi(t)= \frac{d}{dt}\int^{t}_{t_0} X(t,\lambda) f(\lambda) \,d \lambda =X(t,t) f(t)=f(t)$$ but then from the first equation $f(t)=A(t)x+ f(t)$ thus $A(t)x =0$ which is not necessarily true? Where am I wrong?

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1 Answer 1

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Hint: You need to use the product rule to differentiate $$\int^{t}_{t_0} X(t,\lambda) f(\lambda) \,d \lambda.$$ Try writing it as $$\int_{t_{0}}^{t} X(t,\lambda)f(\lambda)\,d\lambda = \int_{{t_{0}}}^{t}X(t)X^{-1}(\lambda)f(\lambda)\,d\lambda=X(t)\int_{t_{0}}^{t}X^{-1}(\lambda)f(\lambda)\,d\lambda.$$

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  • $\begingroup$ So this gives me $\frac{d}{dt}X(t)\int_{t_{0}}^{t}X^{-1}(\lambda)f(\lambda)\,d\lambda= X'(t) \int_{t_{0}}^t X^{-1} (\lambda)f(\lambda) \,d\lambda + X(t)X^{-1}(t)f(t)$ yes ? $\endgroup$
    – huberttt
    Commented Jun 18, 2017 at 22:12
  • $\begingroup$ @huberttt yes, now use the definition of fundamental matrix to rewrite $X'(t)$. $\endgroup$
    – DMcMor
    Commented Jun 18, 2017 at 22:15
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    $\begingroup$ ookay so I put $A(t)X$ instead $X'(t)$ and it works perfectly . Thank you very much . $\endgroup$
    – huberttt
    Commented Jun 18, 2017 at 22:19

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