# Acceleration and Ode's

• Mar 6th 2008, 05:07 PM
reivera
Acceleration and Ode's
ok this prob is a little trickier (x double dot) is given. in gen as a func of x, x dot and t, and here by expression -dot of x + 2e^(-t)+1

Find solution for x(t)
which satisfies x(0)= dot of x(0)=0

I can do init conditions ok on this I think, I'd just like to see how the process of the equation works on this, its probz done in a similar way to the previous one...
• Mar 6th 2008, 08:21 PM
mr fantastic
Quote:

Originally Posted by reivera
ok this prob is a little trickier (x double dot) is given. in gen as a func of x, x dot and t, and here by expression -dot of x + 2e^(-t)+1

Find solution for x(t)
which satisfies x(0)= dot of x(0)=0

I can do init conditions ok on this I think, I'd just like to see how the process of the equation works on this, its probz done in a similar way to the previous one...

Substitute $v = \frac{dx}{dt}$ and solve the first order DE

$\frac{dv}{dt} = -v + 2 e^{-t} + 1 \Rightarrow \frac{dv}{dt} + v = 2 e^{-t} + 1 \,$ subject to the initial condition v(0) = 0. I recommend the integrating factor technique.

Then solve the first order DE

$\frac{dx}{dt} = v(t)\,$ subject to the initial condition x(0) = 0.