D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution.


Example 1

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 1)

Solve the following IVP.

\[{y''} + 4 \, {y} = -4 \, u\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 6\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{6}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{6}{s^{2} + 4}\]

\[{y} = \cos\left(2 \, t - 6\right) u\left(t - 3\right) + 3 \, \sin\left(2 \, t\right) - u\left(t - 3\right)\]


Example 2

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 2)

Solve the following IVP.

\[{y''} + 2 \, {y'} - 3 \, {y} = 16 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 12\]

Hint: \(\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} + 2 \, s - 3} + \frac{12}{s^{2} + 2 \, s - 3}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{3}{s + 3} + \frac{3}{s - 1}\]

\[{y} = 4 \, e^{\left(t - 3\right)} u\left(t - 3\right) - 4 \, e^{\left(-3 \, t + 9\right)} u\left(t - 3\right) - 3 \, e^{\left(-3 \, t\right)} + 3 \, e^{t}\]


Example 3

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 3)

Solve the following IVP.

\[{y''} + {y'} - 12 \, {y} = 14 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= -14\]

Hint: \(\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{14}{s^{2} + s - 12}\]

\[\mathcal L\left\{y\right\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 4} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{2}{s + 4} - \frac{2}{s - 3}\]

\[{y} = 2 \, e^{\left(3 \, t - 9\right)} u\left(t - 3\right) - 2 \, e^{\left(-4 \, t + 12\right)} u\left(t - 3\right) - 2 \, e^{\left(3 \, t\right)} + 2 \, e^{\left(-4 \, t\right)}\]


Example 4

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 4)

Solve the following IVP.

\[{y''} + {y} = -3 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= 1\]

Hint: \(\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{1}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{1}{s^{2} + 1}\]

\[{y} = 3 \, \cos\left(t - 2\right) u\left(t - 2\right) + \sin\left(t\right) - 3 \, u\left(t - 2\right)\]


Example 5

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 5)

Solve the following IVP.

\[{y''} + {y'} - 2 \, {y} = -12 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 6\]

Hint: \(\frac{1}{s^{2} + s - 2} = -\frac{1}{3 \, {\left(s + 2\right)}} + \frac{1}{3 \, {\left(s - 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 2} + \frac{6}{s^{2} + s - 2}\]

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2}{s + 2} + \frac{2}{s - 1}\]

\[{y} = -4 \, e^{\left(t - 3\right)} u\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} u\left(t - 3\right) - 2 \, e^{\left(-2 \, t\right)} + 2 \, e^{t}\]


Example 6

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 6)

Solve the following IVP.

\[{y''} - {y'} - 6 \, {y} = -15 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -20\]

Hint: \(\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{15 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}\]

\[{y} = -3 \, e^{\left(3 \, t - 6\right)} u\left(t - 2\right) + 3 \, e^{\left(-2 \, t + 4\right)} u\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}\]


Example 7

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 7)

Solve the following IVP.

\[{y''} + 9 \, {y} = 36 \, u\left(t - 3\right)\hspace{2em}y(0)= 3 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{3 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}\]

\[{y} = -4 \, \cos\left(3 \, t - 9\right) u\left(t - 3\right) + 3 \, \cos\left(3 \, t\right) + 4 \, u\left(t - 3\right)\]


Example 8

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 8)

Solve the following IVP.

\[{y''} + {y'} - 2 \, {y} = -12 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= -9\]

Hint: \(\frac{1}{s^{2} + s - 2} = -\frac{1}{3 \, {\left(s + 2\right)}} + \frac{1}{3 \, {\left(s - 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 2} - \frac{9}{s^{2} + s - 2}\]

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{3}{s + 2} - \frac{3}{s - 1}\]

\[{y} = -4 \, e^{\left(t - 3\right)} u\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} u\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{t}\]


Example 9

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 9)

Solve the following IVP.

\[{y''} - 5 \, {y'} + 6 \, {y} = 4 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= 4\]

Hint: \(\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}\]

\[{y} = 4 \, e^{\left(3 \, t - 6\right)} u\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} u\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}\]


Example 10

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 10)

Solve the following IVP.

\[{y''} - 16 \, {y} = -32 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -24\]

Hint: \(\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{32 \, e^{\left(-2 \, s\right)}}{s^{2} - 16} - \frac{24}{s^{2} - 16}\]

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s + 4} - \frac{3}{s - 4}\]

\[{y} = -4 \, e^{\left(4 \, t - 8\right)} u\left(t - 2\right) + 4 \, e^{\left(-4 \, t + 8\right)} u\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(-4 \, t\right)}\]


Example 11

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 11)

Solve the following IVP.

\[{y''} - 2 \, {y'} - 8 \, {y} = 24 \, \delta\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 6\]

Hint: \(\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 8} + \frac{6}{s^{2} - 2 \, s - 8}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 4} - \frac{1}{s + 2} + \frac{1}{s - 4}\]

\[{y} = 4 \, e^{\left(4 \, t - 4\right)} u\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} u\left(t - 1\right) + e^{\left(4 \, t\right)} - e^{\left(-2 \, t\right)}\]


Example 12

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 12)

Solve the following IVP.

\[{y''} - 16 \, {y} = -8 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 16\]

Hint: \(\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}\]

\[\mathcal L\left\{y\right\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}\]

\[{y} = -e^{\left(4 \, t - 12\right)} u\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} u\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}\]


Example 13

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 13)

Solve the following IVP.

\[{y''} + 9 \, {y} = -27 \, u\left(t - 1\right)\hspace{2em}y(0)= -3 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{3 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}\]

\[{y} = 3 \, \cos\left(3 \, t - 3\right) u\left(t - 1\right) - 3 \, \cos\left(3 \, t\right) - 3 \, u\left(t - 1\right)\]


Example 14

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 14)

Solve the following IVP.

\[{y''} - 2 \, {y'} - 8 \, {y} = -24 \, \delta\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 24\]

Hint: \(\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{24 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 8} + \frac{24}{s^{2} - 2 \, s - 8}\]

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-s\right)}}{s + 2} - \frac{4 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 2} + \frac{4}{s - 4}\]

\[{y} = -4 \, e^{\left(4 \, t - 4\right)} u\left(t - 1\right) + 4 \, e^{\left(-2 \, t + 2\right)} u\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}\]


Example 15

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 15)

Solve the following IVP.

\[{y''} + 5 \, {y'} + 6 \, {y} = 2 \, \delta\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= -2\]

Hint: \(\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{2}{s^{2} + 5 \, s + 6}\]

\[\mathcal L\left\{y\right\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2}{s + 3} - \frac{2}{s + 2}\]

\[{y} = 2 \, e^{\left(-2 \, t + 2\right)} u\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} u\left(t - 1\right) - 2 \, e^{\left(-2 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}\]


Example 16

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 16)

Solve the following IVP.

\[{y''} + 4 \, {y} = -16 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -2\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{2}{s^{2} + 4} - \frac{16 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{4 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s^{2} + 4}\]

\[{y} = 4 \, \cos\left(2 \, t - 4\right) u\left(t - 2\right) - \sin\left(2 \, t\right) - 4 \, u\left(t - 2\right)\]


Example 17

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 17)

Solve the following IVP.

\[{y''} - 3 \, {y'} - 4 \, {y} = -20 \, \delta\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= -20\]

Hint: \(\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{20 \, e^{\left(-s\right)}}{s^{2} - 3 \, s - 4} - \frac{20}{s^{2} - 3 \, s - 4}\]

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4 \, e^{\left(-s\right)}}{s - 4} + \frac{4}{s + 1} - \frac{4}{s - 4}\]

\[{y} = -4 \, e^{\left(4 \, t - 4\right)} u\left(t - 1\right) + 4 \, e^{\left(-t + 1\right)} u\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-t\right)}\]


Example 18

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 18)

Solve the following IVP.

\[{y''} + 7 \, {y'} + 12 \, {y} = -3 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= 1\]

Hint: \(\frac{1}{s^{2} + 7 \, s + 12} = -\frac{1}{s + 4} + \frac{1}{s + 3}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 7 \, s + 12} + \frac{1}{s^{2} + 7 \, s + 12}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{1}{s + 4} + \frac{1}{s + 3}\]

\[{y} = -3 \, e^{\left(-3 \, t + 6\right)} u\left(t - 2\right) + 3 \, e^{\left(-4 \, t + 8\right)} u\left(t - 2\right) + e^{\left(-3 \, t\right)} - e^{\left(-4 \, t\right)}\]


Example 19

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 19)

Solve the following IVP.

\[{y''} + 9 \, {y} = 27 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -6\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{6}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{6}{s^{2} + 9}\]

\[{y} = -3 \, \cos\left(3 \, t - 6\right) u\left(t - 2\right) - 2 \, \sin\left(3 \, t\right) + 3 \, u\left(t - 2\right)\]


Example 20

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 20)

Solve the following IVP.

\[{y''} + 4 \, {y} = 16 \, u\left(t - 2\right)\hspace{2em}y(0)= -5 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{5 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-2 \, s\right)}}{s}\]

\[{y} = -4 \, \cos\left(2 \, t - 4\right) u\left(t - 2\right) - 5 \, \cos\left(2 \, t\right) + 4 \, u\left(t - 2\right)\]


Example 21

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 21)

Solve the following IVP.

\[{y''} + {y} = -3 \, u\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 2\]

Hint: \(\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{2}{s^{2} + 1} - \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{2}{s^{2} + 1}\]

\[{y} = 3 \, \cos\left(t - 1\right) u\left(t - 1\right) + 2 \, \sin\left(t\right) - 3 \, u\left(t - 1\right)\]


Example 22

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 22)

Solve the following IVP.

\[{y''} + 4 \, {y} = -4 \, u\left(t - 1\right)\hspace{2em}y(0)= 4 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{4 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{s e^{\left(-s\right)}}{s^{2} + 4} + \frac{4 \, s}{s^{2} + 4} - \frac{e^{\left(-s\right)}}{s}\]

\[{y} = \cos\left(2 \, t - 2\right) u\left(t - 1\right) + 4 \, \cos\left(2 \, t\right) - u\left(t - 1\right)\]


Example 23

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 23)

Solve the following IVP.

\[{y''} + 5 \, {y'} + 4 \, {y} = 3 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 6\]

Hint: \(\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 4} + \frac{6}{s^{2} + 5 \, s + 4}\]

\[\mathcal L\left\{y\right\}= -\frac{e^{\left(-3 \, s\right)}}{s + 4} + \frac{e^{\left(-3 \, s\right)}}{s + 1} - \frac{2}{s + 4} + \frac{2}{s + 1}\]

\[{y} = e^{\left(-t + 3\right)} u\left(t - 3\right) - e^{\left(-4 \, t + 12\right)} u\left(t - 3\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-4 \, t\right)}\]


Example 24

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 24)

Solve the following IVP.

\[{y''} - 7 \, {y'} + 12 \, {y} = -\delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 4\]

Hint: \(\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{4}{s^{2} - 7 \, s + 12}\]

\[\mathcal L\left\{y\right\}= \frac{e^{\left(-3 \, s\right)}}{s - 3} - \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s - 3} + \frac{4}{s - 4}\]

\[{y} = -e^{\left(4 \, t - 12\right)} u\left(t - 3\right) + e^{\left(3 \, t - 9\right)} u\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(3 \, t\right)}\]


Example 25

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 25)

Solve the following IVP.

\[{y''} + 9 \, {y} = 27 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -12\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{12}{s^{2} + 9}\]

\[{y} = -3 \, \cos\left(3 \, t - 6\right) u\left(t - 2\right) - 4 \, \sin\left(3 \, t\right) + 3 \, u\left(t - 2\right)\]


Example 26

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 26)

Solve the following IVP.

\[{y''} + 9 \, {y} = -27 \, u\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 12\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{12}{s^{2} + 9}\]

\[{y} = 3 \, \cos\left(3 \, t - 9\right) u\left(t - 3\right) + 4 \, \sin\left(3 \, t\right) - 3 \, u\left(t - 3\right)\]


Example 27

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 27)

Solve the following IVP.

\[{y''} - {y'} - 6 \, {y} = -5 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -5\]

Hint: \(\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{5 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 6} - \frac{5}{s^{2} - s - 6}\]

\[\mathcal L\left\{y\right\}= \frac{e^{\left(-2 \, s\right)}}{s + 2} - \frac{e^{\left(-2 \, s\right)}}{s - 3} + \frac{1}{s + 2} - \frac{1}{s - 3}\]

\[{y} = -e^{\left(3 \, t - 6\right)} u\left(t - 2\right) + e^{\left(-2 \, t + 4\right)} u\left(t - 2\right) - e^{\left(3 \, t\right)} + e^{\left(-2 \, t\right)}\]


Example 28

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 28)

Solve the following IVP.

\[{y''} + {y'} - 2 \, {y} = 6 \, \delta\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= -12\]

Hint: \(\frac{1}{s^{2} + s - 2} = -\frac{1}{3 \, {\left(s + 2\right)}} + \frac{1}{3 \, {\left(s - 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} + s - 2} - \frac{12}{s^{2} + s - 2}\]

\[\mathcal L\left\{y\right\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s - 1} + \frac{4}{s + 2} - \frac{4}{s - 1}\]

\[{y} = 2 \, e^{\left(t - 1\right)} u\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} u\left(t - 1\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{t}\]


Example 29

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 29)

Solve the following IVP.

\[{y''} - {y'} - 12 \, {y} = 7 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= 7\]

Hint: \(\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} + \frac{7}{s^{2} - s - 12}\]

\[\mathcal L\left\{y\right\}= -\frac{e^{\left(-3 \, s\right)}}{s + 3} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s + 3} + \frac{1}{s - 4}\]

\[{y} = e^{\left(4 \, t - 12\right)} u\left(t - 3\right) - e^{\left(-3 \, t + 9\right)} u\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(-3 \, t\right)}\]


Example 30

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 30)

Solve the following IVP.

\[{y''} + 9 \, {y} = -27 \, u\left(t - 2\right)\hspace{2em}y(0)= 1 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} + \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}\]

\[{y} = 3 \, \cos\left(3 \, t - 6\right) u\left(t - 2\right) + \cos\left(3 \, t\right) - 3 \, u\left(t - 2\right)\]


Example 31

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 31)

Solve the following IVP.

\[{y''} + 4 \, {y} = 4 \, u\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 4\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{4}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s} + \frac{4}{s^{2} + 4}\]

\[{y} = -\cos\left(2 \, t - 2\right) u\left(t - 1\right) + 2 \, \sin\left(2 \, t\right) + u\left(t - 1\right)\]


Example 32

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 32)

Solve the following IVP.

\[{y''} + 4 \, {y} = 16 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -8\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{8}{s^{2} + 4} + \frac{16 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}\]

\[{y} = -4 \, \cos\left(2 \, t - 4\right) u\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 4 \, u\left(t - 2\right)\]


Example 33

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 33)

Solve the following IVP.

\[{y''} - 2 \, {y'} - 8 \, {y} = 18 \, \delta\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= -12\]

Hint: \(\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{18 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 8} - \frac{12}{s^{2} - 2 \, s - 8}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, e^{\left(-s\right)}}{s + 2} + \frac{3 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 2} - \frac{2}{s - 4}\]

\[{y} = 3 \, e^{\left(4 \, t - 4\right)} u\left(t - 1\right) - 3 \, e^{\left(-2 \, t + 2\right)} u\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}\]


Example 34

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 34)

Solve the following IVP.

\[{y''} + 9 \, {y} = -18 \, u\left(t - 2\right)\hspace{2em}y(0)= -5 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{5 \, s}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{5 \, s}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s}\]

\[{y} = 2 \, \cos\left(3 \, t - 6\right) u\left(t - 2\right) - 5 \, \cos\left(3 \, t\right) - 2 \, u\left(t - 2\right)\]


Example 35

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 35)

Solve the following IVP.

\[{y''} + 9 \, {y} = 36 \, u\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 9\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{9}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s} + \frac{9}{s^{2} + 9}\]

\[{y} = -4 \, \cos\left(3 \, t - 3\right) u\left(t - 1\right) + 3 \, \sin\left(3 \, t\right) + 4 \, u\left(t - 1\right)\]


Example 36

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 36)

Solve the following IVP.

\[{y''} + 4 \, {y} = 4 \, u\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= -2\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{2}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} + \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 4}\]

\[{y} = -\cos\left(2 \, t - 6\right) u\left(t - 3\right) - \sin\left(2 \, t\right) + u\left(t - 3\right)\]


Example 37

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 37)

Solve the following IVP.

\[{y''} - 5 \, {y'} + 6 \, {y} = 4 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -3\]

Hint: \(\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}\]

\[{y} = 4 \, e^{\left(3 \, t - 6\right)} u\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} u\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}\]


Example 38

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 38)

Solve the following IVP.

\[{y''} + {y} = -3 \, u\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= -4\]

Hint: \(\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{4}{s^{2} + 1} - \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s^{2} + 1}\]

\[{y} = 3 \, \cos\left(t - 1\right) u\left(t - 1\right) - 4 \, \sin\left(t\right) - 3 \, u\left(t - 1\right)\]


Example 39

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 39)

Solve the following IVP.

\[{y''} + {y} = 2 \, u\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= -1\]

Hint: \(\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{1}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{1}{s^{2} + 1}\]

\[{y} = -2 \, \cos\left(t - 3\right) u\left(t - 3\right) - \sin\left(t\right) + 2 \, u\left(t - 3\right)\]


Example 40

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 40)

Solve the following IVP.

\[{y''} - 9 \, {y} = 24 \, \delta\left(t - 3\right)\hspace{2em}y(0)= 0 ,y'(0)= -6\]

Hint: \(\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{24 \, e^{\left(-3 \, s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}\]

\[{y} = 4 \, e^{\left(3 \, t - 9\right)} u\left(t - 3\right) - 4 \, e^{\left(-3 \, t + 9\right)} u\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}\]


Example 41

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 41)

Solve the following IVP.

\[{y''} + 9 \, {y} = 9 \, u\left(t - 1\right)\hspace{2em}y(0)= 5 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{5 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 9} + \frac{5 \, s}{s^{2} + 9} + \frac{e^{\left(-s\right)}}{s}\]

\[{y} = -\cos\left(3 \, t - 3\right) u\left(t - 1\right) + 5 \, \cos\left(3 \, t\right) + u\left(t - 1\right)\]


Example 42

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 42)

Solve the following IVP.

\[{y''} + 9 \, {y} = 27 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= -6\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{6}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{6}{s^{2} + 9}\]

\[{y} = -3 \, \cos\left(3 \, t - 6\right) u\left(t - 2\right) - 2 \, \sin\left(3 \, t\right) + 3 \, u\left(t - 2\right)\]


Example 43

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 43)

Solve the following IVP.

\[{y''} + 9 \, {y} = -27 \, u\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 12\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}\]

\[{y} = 3 \, \cos\left(3 \, t - 3\right) u\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, u\left(t - 1\right)\]


Example 44

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 44)

Solve the following IVP.

\[{y''} - {y} = 6 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= 4\]

Hint: \(\frac{1}{s^{2} - 1} = -\frac{1}{2 \, {\left(s + 1\right)}} + \frac{1}{2 \, {\left(s - 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 1} + \frac{4}{s^{2} - 1}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 1} - \frac{2}{s + 1} + \frac{2}{s - 1}\]

\[{y} = 3 \, e^{\left(t - 2\right)} u\left(t - 2\right) - 3 \, e^{\left(-t + 2\right)} u\left(t - 2\right) - 2 \, e^{\left(-t\right)} + 2 \, e^{t}\]


Example 45

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 45)

Solve the following IVP.

\[{y''} + 4 \, {y} = -12 \, u\left(t - 3\right)\hspace{2em}y(0)= -4 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-3 \, s\right)}}{s}\]

\[{y} = 3 \, \cos\left(2 \, t - 6\right) u\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) - 3 \, u\left(t - 3\right)\]


Example 46

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 46)

Solve the following IVP.

\[{y''} + 9 \, {y} = -18 \, u\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= 3\]

Hint: \(\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}\]

\[{y} = 2 \, \cos\left(3 \, t - 6\right) u\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, u\left(t - 2\right)\]


Example 47

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 47)

Solve the following IVP.

\[{y''} + 5 \, {y'} + 4 \, {y} = 9 \, \delta\left(t - 2\right)\hspace{2em}y(0)= 0 ,y'(0)= 9\]

Hint: \(\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{9 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 4} + \frac{9}{s^{2} + 5 \, s + 4}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s + 1} - \frac{3}{s + 4} + \frac{3}{s + 1}\]

\[{y} = 3 \, e^{\left(-t + 2\right)} u\left(t - 2\right) - 3 \, e^{\left(-4 \, t + 8\right)} u\left(t - 2\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-4 \, t\right)}\]


Example 48

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 48)

Solve the following IVP.

\[{y''} + 4 \, {y} = 16 \, u\left(t - 1\right)\hspace{2em}y(0)= 0 ,y'(0)= 6\]

Hint: \(\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{6}{s^{2} + 4} + \frac{16 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 4}\]

\[{y} = -4 \, \cos\left(2 \, t - 2\right) u\left(t - 1\right) + 3 \, \sin\left(2 \, t\right) + 4 \, u\left(t - 1\right)\]


Example 49

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 49)

Solve the following IVP.

\[{y''} + {y} = -2 \, u\left(t - 2\right)\hspace{2em}y(0)= 2 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}\).

Answer.

\[\mathcal L\left\{y\right\}= \frac{2 \, s}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}\]

\[\mathcal L\left\{y\right\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{2 \, s}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s}\]

\[{y} = 2 \, \cos\left(t - 2\right) u\left(t - 2\right) + 2 \, \cos\left(t\right) - 2 \, u\left(t - 2\right)\]


Example 50

D2 - Discontinuous IVPs. Use Laplace transforms to solve IVPs involving a step function or Dirac delta distribution. (ver. 50)

Solve the following IVP.

\[{y''} + {y} = 3 \, u\left(t - 2\right)\hspace{2em}y(0)= -4 ,y'(0)= 0\]

Hint: \(\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}\).

Answer.

\[\mathcal L\left\{y\right\}= -\frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}\]

\[\mathcal L\left\{y\right\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}\]

\[{y} = -3 \, \cos\left(t - 2\right) u\left(t - 2\right) - 4 \, \cos\left(t\right) + 3 \, u\left(t - 2\right)\]