Solve $y'' -4y' + 3y = 2cos(at) + \delta(t-1) , y(0) = 0, y'(0) = 1$ where $a>0$.

Question

Applying Laplace transforms gives:

$(s^2 -4s + 3)y(s) -1 = \frac{2s}{s^2 + a^2} + e^{-s}$

$\Rightarrow y(s) = \frac{2s}{(s-3)(s-1)(s^2 + a^2)} + \frac{e^{-s}}{(s-3)(s-1)} + \frac{1}{(s-3)(s-1)} $

After partial fractions we then get:

$y(s) = \frac{3}{(9+a^2)(s-3)} - \frac{1}{(1+a^2)(s-1)} - \frac{8a^2}{(9+a^2)(1+a^2)(s^2+a^2)} + e^{-s}(\frac{1}{2(s-3)}) - e^{-s}(\frac{1}{2(s-1)}) + \frac{1}{2(s-3)} - \frac{1}{2(s-1)}$

Inverse Laplace each term:

$\frac{3}{9+a^2}\mathcal{L}^{-1}(\frac{1}{s-3}) - \frac{1}{1+a^2}\mathcal{L}^{-1}(\frac{1}{s-1})- \frac{8a^2}{(9+a^2)(1+a^2)}\mathcal{L}^{-1}(\frac{1}{s^2 + a^2}) + \frac{1}{2}\mathcal{L}^{-1}(e^{-s}\frac{1}{s-3}) - \frac{1}{2}\mathcal{L}^{-1}(e^{-s}\frac{1}{s-1}) + \frac{1}{2}\mathcal{L}^{-1}(\frac{1}{s-3}) - \frac{1}{2}\mathcal{L}^{-1}(\frac{1}{s-1}) $

Which gives:

$ y = \frac{3e^{3t}}{9+a^2} - \frac{e^t}{1+a^2} - \frac{8a^2sin(at)}{(9+a^2)(1+a^2)} + \frac{e^{3t}}{2} - \frac{e^{t}}{2} + \frac{1}{2}H(t-1)e^{3t} + \frac{1}{2}H(t-1)e^{t} $

where $H(t-\alpha)$ is the Heaviside function.

This is Wolfram's solution: http://www.wolframalpha.com/input/?i=y%27%27(t)+-+4y%27(t)+%2B3y(t)+%3D+2cos(wt)+%2B+delta(t-1)

Which is different to mine...does anyone know what I have done wrong?

Answer 1

Here is what wolfram alpha gave me.

Answer 2

Sometimes the answer to a differential equation may be represented in a different format but they actually represent the same thing. Moreover they may not be unique. Try validating your answer by plugging in the $y(s)$ in the differential equation provided.