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I have been given a variation of parameters formula for a first order ODE $x'(t) = a(t)x(t)+b(t)$ and have been asked to differentiate it, the formula is: $$ x(t) = Ce^{\int_{t_0}^{t}a(s)\,ds} + \int_{t_0}^{t}b(u)e^{\int_{u}^{t}a(s)\,ds}\,du $$

The solutions say that an intermediate step is: $$ x'(t) = Ca(t)e^{\int_{t_0}^{t}a(s)\,ds} + b(t) +\int_{t_0}^{t}b(u)a(t)e^{\int_{u}^{t}a(s)\,ds}\,du $$

Would it be possible to explain this intermediate step to me?

Note by $x'(t)$ I mean the derivative of $x$ with respect to $t$.

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  • $\begingroup$ Is the derivative of the first term clear? The second term differentiation is essentially by the product rule. Note that you can extract $$e^{\int_{t_0}^{t}a(s)\,ds}$$ from the second integral so that the product structure becomes explicit. $\endgroup$ – LutzL Apr 22 at 11:47
  • $\begingroup$ Ahh thank you so much! Yeah the first term is clear I didn't spot the trick to pull out the integral you mentioned. Thanks! $\endgroup$ – Milos Tasic Apr 22 at 11:56

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