# variation of parameters 1st order ODE Version

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 $$t$$.

• 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. – LutzL Apr 22 at 11:47
• 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! – Milos Tasic Apr 22 at 11:56