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Solution to differential equation
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[QUOTE="Simon Bridge, post: 5000335, member: 367532"] Puzzling, the chain rule is usually taught in year 1 calculus classes, right after the product and quotient rules. Are you self-taught? $$\frac{dv}{dt} = \frac{d\tau}{dt}\frac{dv}{d\tau}$$ Changing variable in the DE by the chain rule: $$\frac{dv}{dt} = \gamma v + \frac{1}{m}F'(t) \implies N\frac{dv}{d\tau} = \gamma v + \frac{1}{m}F'(\tau): \tau=Nt$$ The proposed solution: $$v-v_0e^{-\gamma\tau} = \Sigma \cdots$$... too lazy to write out the whole sum... but taking the derivative of both sides: $$\begin{aligned} N\frac{dv}{d \tau} + v_0\gamma e^{-\gamma \tau} &= \sum_{k=0}^{N-1} \left[ e^{\gamma \tau(N-k)}\frac{d}{d \tau}\int_0^\tau F'(k\tau+s)\;\text{d}s \;\; -\gamma(N-k)e^{-\gamma\tau(N-k)}\int_0^\tau F'(k\tau+s)\;\text{d}s \right] \\ & = N\gamma v + \frac{N}{m}F'(\tau) + v_0\gamma e^{-\gamma\tau} \end{aligned} $$ ... something like that. Check my work. So the summation (above) has to come out to the last line. 1st you want to figure out if you can take the derivative inside the integration. i.e. can you say: $$\frac{d}{d\tau}\int_0^\tau F'(k\tau+s)\;\text{d}s = \int_0^\tau \left(\frac{d}{d\tau}F'(k\tau+s)\right)\;\text{d}s$$ ... if you can, you can use the chain rule on the integrand, which will go some way to simplifying the calculation. Can you simplify further by, say, cancelling terms? The way to get rid of the summation, usually, is either to show that it converges to the part that is not the sum, or see if you can make the sum here look like the RHS of the proposed solution. I don't see it being nice. The other approach is just to solve the E yourself and show that the solution you get it the same as theirs. [/QUOTE]
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Solution to differential equation
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