# Math Help - Different Results when Using Laplace or Complamentary Functions/Particular Integrals

1. ## Different Results when Using Laplace or Complamentary Functions/Particular Integrals

Hi,

y''(t) + 3y'(t) + y(t) = 2x(t)

Where x(t) = u(t), and the system is initially at rest. We are interested only when t>=0, so x(t) = 1 always.

I can solve it by Laplace transforms:

Original eqn:
y''(t) + 3y'(t) + y(t) = 2

Taking Laplace of both sides (and assuming system was at rest):
Y(s)[s^2 + 3s + 2] = 2/s
Y(s) = [2/s]/[s^2 + 3s + 2]
Y(s) = A/s + B/(s + 1) + C/(s + 2)

Finally: y(t) = 1 - 2exp[-t] + ext[-2t]

--------

The problem is when I try solving it by equating coefficients. I get a different answer:

Original eqn:
y''(t) + 3y'(t) + y(t) = 2

To get complementary function, y_comp(t):
y''(t) + 3y'(t) + y(t) = 0
Auxiliary eqn: s^2 + 3s + 1 = 0; s = -1, -2
y_comp(t) = Aexp[-t] + Bexp[-2t]

To get particular integral, y_pt(t):
Forcing function: x(t) = 1
Gives y_pt(t) = 1

General solution:
y(t) = y_comp(t) + y_pt(t) = Aexp[-t] + Bexp[-2t] + 1

Solving for the constants A and B (problem is here):
y'(t) = -Aexp[-t] -2Bexp[-2t]
y''(t) = Aexp[-t] + 4Bexp[-2t]
Given the system was at rest, I get:
0 = -A - 2B (1)
0 = A + 4B (2)

Solving these gives A=B=0. But this means that y(t) = 1, which is different to the Laplace solution of y(t) = 1 - 2exp[-t] + ext[-2t].

Thanks for any help.

2. Originally Posted by algorithm
Hi,

y''(t) + 3y'(t) + y(t) = 2x(t)

Where x(t) = u(t), and the system is initially at rest. We are interested only when t>=0, so x(t) = 1 always.

I can solve it by Laplace transforms:

Original eqn:
y''(t) + 3y'(t) + y(t) = 2

Taking Laplace of both sides (and assuming system was at rest):
Y(s)[s^2 + 3s + 2] = 2/s
Y(s) = [2/s]/[s^2 + 3s + 2]
Y(s) = A/s + B/(s + 1) + C/(s + 2)

Finally: y(t) = 1 - 2exp[-t] + ext[-2t]

--------

The problem is when I try solving it by equating coefficients. I get a different answer:

Original eqn:
y''(t) + 3y'(t) + y(t) = 2

To get complementary function, y_comp(t):
y''(t) + 3y'(t) + y(t) = 0
Auxiliary eqn: s^2 + 3s + 1 = 0; s = -1, -2
y_comp(t) = Aexp[-t] + Bexp[-2t]

To get particular integral, y_pt(t):
Forcing function: x(t) = 1
Gives y_pt(t) = 1

General solution:
y(t) = y_comp(t) + y_pt(t) = Aexp[-t] + Bexp[-2t] + 1

Solving for the constants A and B (problem is here):
y'(t) = -Aexp[-t] -2Bexp[-2t]
y''(t) = Aexp[-t] + 4Bexp[-2t]
Given the system was at rest, I get:
0 = -A - 2B (1)
0 = A + 4B (2)

Solving these gives A=B=0. But this means that y(t) = 1, which is different to the Laplace solution of y(t) = 1 - 2exp[-t] + ext[-2t].

Thanks for any help.
When you apply initial conditions, there isn't one for the second derivative of the function!

$A+B+1=0$ (for $y(0)=0$)

$-A-2B=0$ (for $y^{\prime}(0)=0$)

Can you take it from here?

3. Hi,

Thanks for the reply. So when we assume a 2nd order system is initially at rest, the initial conditions are only y(0) = y'(0) = 0?

Thanks

4. The second order linear DE is...

$y^{''} (t) + 3\cdot y^{'} (t) + 2\cdot y(t) = 2\cdot \mathcal{U} (t)$ , $y(0)=y^{'}(0)=0$ (1)

... being $\mathcal{U} (*)$ the 'Haeviside Step Function'. In terms of LT the (1) becomes...

$s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) +3\cdot s\cdot Y(s) -3\cdot y(0) + 2\cdot Y(s) = \frac{2}{s}$ (2)

... and from it...

$Y(s)= \frac{(s-3)\cdot y(0) + y^{'} (0)}{s^{2} + 3s +2} + \frac{2}{s\cdot (s^{2} + 3s + 2)}$ (3)

Because is $y(0)=y^{'}(0)=0$ the only contribution is the 'forced response' ...

$Y(s) = \frac{2}{s\cdot (s^{2} + 3s + 2)} = \frac{1}{s} - \frac{2}{s+1} + \frac{1}{s+2}$ (4)

... so that is...

$y(t)=\mathcal{L}^{-1} \{Y(s)\} = (1-2\cdot e^{-t} + e^{-2t})\cdot \mathcal{U} (t)$ (5)

Kind regards

$\chi$ $\sigma$