Question

I am trying to solve a second order non-homogeneous differential equation where x(t) has u(t), the unit step as a part. i.e. x(t)=f(t)u(t)

I know how to 'guess' the particular solution for f(t) alone, but I'm not sure how it should be done when the unit step is appended.

Is there a standard method for guessing the particular solution in this scenario?

e.g. x(t)= Sin(3t)u(t), e^(-t)u(t),...

Answer 1

what's your original equation?

Answer 2

Take
\[y \prime \prime + 2y \prime +2 = e^{-t} u(t)\]
as an example where y is a function of t.

Answer 3

is u(t) step function?

Answer 4

Pretty sure this is easily solvable using the Laplace transform.$$y''+2y'+2=e^{-t}\\\mathcal{L}\{y''+2y'+2\}=\mathcal{L}\{e^{-t}u(t)\}\\s^2Y-sy(0)-y'(0)+2sY-2y(0)+\frac2s=\frac1{s+1}\\s(s+2)Y-sy(0)-y'(0)-2y(0)=\frac1{s+1}-\frac2s=-\frac{s+2}{s(s+1)}\\s(s+2)Y=\frac1{s+1}-\frac2s+sy(0)+y'(0)+2y(0)\\(s+2)Y=\frac1{s(s+1)}-\frac2{s^2}+y(0)+\frac{y'(0)}s+\frac{2y'(0)}s\\Y=\frac1{s(s+1)(s+2)}-\frac2{s^2(s+2)}+\frac{y(0)}{s+2}+\frac{y'(0)}{s(s+2)}+\frac{2y'(0)}{s(s+2)}\\Y=-\frac1{s^2}+\frac1s-\frac1{s+1}+\frac{y(0)}s+\frac{y'(0)}{2s}-\frac{y'(0)}{2(s+2)}\\y=-t-\cosh t+\sinh t+y(0)+\frac12y'(0)-\frac12y'(0)e^{2t}+1$$

Answer 5

consider two cases ... before t>0 and t<0, you will have two solutions for two different inputs. the solution should be piece-wise function of t