How to solve x"+2.55x'+3.81x = f(t) where f(t) = 0.51sin(t)+1.7+0.4t for 0<=t<=2

Question

The remaining question is Linear f(t): (t2=0, f2=0), (t3=3.05 f3=1.63), (t4=4.15 f4=1.11), (t5=5 f5=0).

Solve the above ODE using Runge Kutta 4th order method

Answer 1

Assume x=e^At: x'=Ae^At, x''=A^2e^At, where A is a constant to be found.

Substitute in the equation x"+2.55x'+3.81x=0:

A^2e^At+2.55Ae^At+3.81e^At=0 so, since e^At can never be zero,

A^2+2.55A+3.81=0 and A=(-2.55±sqrt(-8.7375))/2=-1.275±isqrt(8.7375)/2=a+ib where a=-1.275 and b=sqrt(8.7375)/2. Using a and b instead of their values simplifies the presentation of the solution. Remember that the sum of the roots is 2.55 and their product is 3.81, i.e., 2a=-2.55 and a^2+b^2=3.81.

So x=e^(a+ib)t or e^(a-ib)t=e^at(e^ibt) or e^at(e^-ibt).

Let x=e^(a+ib)t+e^(a-ib)t because both complex roots are valid and the differential (homogeneous) part of the solution applies to both roots.

But e^ip=cos(p)+isin(p), so x=e^at(cos(bt)+isin(bt))+e^at(cos(bt)-isin(bt))=2e^atcos(bt).

We need to check so far: x=2e^(at)cos(bt),

x'=-2be^(at)sin(bt)+2ae^(at)cos(bt),

x"=-2b^2e^(at)cos(bt)-2abe^(at)sin(bt)-2abe^(at)sin(bt)+2a^2e^(at)cos(bt).

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