ODE RC Circuit Problem

Homework Statement

Suppose an RC series circuit has a variable resistor. If the resistance at time t is given by by R = a + bt, where a and b are known positive constants then the charge q(t) on the capacitor satisfies

(a+bt) q' + (1/C)q = V

where V is some constant. Also q(0) = q_0
Find q(t) as an explicit function of t.

Homework Equations

Now I have obtained the answer, however my main question is: am I allowed to treat C (capacitance) as a constant in this equation. It doesn't specify in the question, but to my knowledge (unless I am wrong of course) capacitance is a constant value and only depends on material and physical parameters of the capacitor itself (how it is build).

If I can't treat C as a constant then I believe there is no way to evaluate the integral in integrating factor and I'd have to leave it as it is.

Process is simple from there, I rewrite the equation in standart form, find the integrating factor and obtain a formula for q(t), evaluate an integration constant with q(0) = q_0 and obtain the overall solution q(t).

The answer looks quite frightening btw

3. My solution:
attached pdf file

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The capacitance must be constant during the all process, which just depends on its material and geometric shape, as what you say above,

vela
Staff Emeritus
Homework Helper
Is there some reason you're not simplifying $\frac{(a+bt)^k}{a+bt}$ to $(a+bt)^{k-1}$ and integrating the righthand side? Also, the $t$ in the integral is a dummy variable, so it's not correct to set it to 0.

That's what my prof does. By setting it from 0 to t, we ensure evaluated integral is equal to the true value with t as a variable, how would setting it to something arbitary as t_o (I'm assuming that's what you mean)be any better? My understanding of it is that we need to choose some "convenient" interval, what's wrong with 0 to t?

And thank you for pointing out I can actually simplify it further, I completely missed that.

vela
Staff Emeritus
Homework Helper
$$\int (a+bt)^2\,dt = \int a\,dt$$ by claiming you're setting $t=0$. The $t$ inside the integral isn't the same $t$ that appears elsewhere.
$$\int (a+bt)^2\,dt = \int a\,dt$$ by claiming you're setting $t=0$. The $t$ inside the integral isn't the same $t$ that appears elsewhere.