Differential Equations

so i have a funtion u(t) where u"+bu'=a.  b and a are both constants.  is this even solvable?  its for my physics 325 class and i'm drawing a complete blank.

First solve u" + bu' = 0
characteristic equation: r^2 + r = 0
So, r = 0 or r = -1
The complementary solution is therefore:
y_c = C_1 e^(-t) + C_2
Since a is a constant, this equation will suffice. Fill in the boundary values and solve C_1 and C_2.

I believe you use the Method of Undetermined Coefficients to solve this one

>>2

You forgot b in this problem.

thank you guys you rock.  just want to make sure im right, but the second part i add (i completely forgot terms, but the methods are coming back) is u=at/b right? so overall the answer is u=C_1 e^(-bt)+ C_2 + at/b?

Formulaic solutions to DEs are fine and dandy but you should be able to solve this by integrating.

Since the DE is ordinary you can reduce the order by making a V = du/dt substitution to get dV/dt +bV =a, which is a separable first order DE.

Some simple re-arranging yields dV/(a-bV) = dt => -(1/b)lnV = t + C1 => V = exp(-bt + C1) (as C1 absorbs the -b) => du/dt = exp(C1)*exp(-bt), we'll say exp(C1) = D1 so we get du/dt = D1*exp(-bt) => du = D1*exp(-bt)dt => u(t) = D1*exp(-bt)/(-b) + D2

We can again capture the -b inside D1 so our solution is

u(t) = D1*exp(-bt) + D2

Which is entirely consistent with >>2