Differential equation current flow help please?

Question

the differential equation governing current flow i(t) in a circuit is given by di/dt + 2i = t. Use the integrating factor method to find i(t) given that i(0) = 0.

Been stuck on this one for a while. Thanks in advance.

Answer 1

you're anti-differentiating this one (taking the integral). I don't know what the "integrating factor method" is. but when you anti-diff, you get a constant C. then use the condition they gave you (i(0) = 0) to find C

Answer 2

Hey mate,

The above equation is known as a first order linear ODE, which in your circumstance can be solved using the integrating factor,

By definition for a ODE of the form

dy/dt + p(t)y(t) = r(t)

The integrating factor is obtained using

IF = exp( integrate(p(t) dt))

in your situation p(t) = 2

Thus,

IF = exp(integrate(2,dt)) = exp(2t)

The next step is to multiply the both sides by the integrating factor,

Thus,

exp(2t) (di/dt + 2i) = exp(2t)*t

Whereby employing the product allows us to reduce the LHS to,

d/dt (exp(2t)*i(t)) = t*exp(2*t)

Integrate both sides w.r.t to yield

int ( d/dt (exp(2t)*i(t) ) = int(t*exp(2*t))

note for the LHS, we are integrating the derivate which yields the original function at hand,

Thus,

exp(2t)*i(t) = int(t*exp(2*t))

for the RHS employ integration by parts

int(uv') = uv - int(vu')

let u = t, v' = exp(2t) --> u' = 1 v = (1/2)exp(2t)

Thus the LHS becomes

int(t*exp(2*t)) = t*exp(2t) - (1/2)int(exp(2*t)

= t*exp(2*t) - (1/4) exp(2t) + k, where k is merely a constant of integration

Therefore we we equation LHS to the RHS we obtain

exp(2*t)*i(t) = (1/2)*t*exp(2t) - (1/4)exp(2t) + k

-->

i(t) = (1/2)*t - (1/4) + kexp(-2*t)

to determine k, we sub in your initial condition i(0) = 0,

Thus

i(0) = 0 = 0 - (1/4) +kexp(0)

0 = -1/4 + k, --> k = 1/4,

Therfore,

i(t) = (1/2)*t - (1/4) + (1/4)exp(-2*t)

Hope this helps and I didnt go overboard with the explanation,

Regards,

David