D4 - Using Laplace transforms to solve IVPs


Example 1 πŸ”—

Explain how to solve the following IVP.

\[ -9 \, {y} + 12 \, {y'} + 24 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 4 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 4 \, s + 3} + \frac{4}{s^{2} - 4 \, s + 3} \]

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

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


Example 2 πŸ”—

Explain how to solve the following IVP.

\[ 2 \, {y''} - 24 \, \delta\left(t - 2\right) = 10 \, {y'} - 8 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -3 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 4} - \frac{3}{s^{2} - 5 \, s + 4} \]

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

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


Example 3 πŸ”—

Explain how to solve the following IVP.

\[ -16 \, {y} - 2 \, {y''} - 12 \, {y'} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2 \]

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

Answer:

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

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

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


Example 4 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 5 πŸ”—

Explain how to solve the following IVP.

\[ -2 \, {y''} = 18 \, {y} - 72 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3 \]

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

Answer:

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

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

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


Example 6 πŸ”—

Explain how to solve the following IVP.

\[ 3 \, {y''} - 6 \, {y'} - 24 \, {y} - 54 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 24 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{18 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{24}{s^{2} - 2 \, s - 8} \]

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

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


Example 7 πŸ”—

Explain how to solve the following IVP.

\[ 3 \, \delta\left(t - 1\right) = 36 \, {y} + 3 \, {y''} - 21 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2 \]

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

Answer:

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

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

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


Example 8 πŸ”—

Explain how to solve the following IVP.

\[ -3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} - 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6 \]

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

Answer:

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

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

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


Example 9 πŸ”—

Explain how to solve the following IVP.

\[ 0 = -18 \, {y} - 2 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{18 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s} \]

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

\[ {y} = -2 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 3\right) \]


Example 10 πŸ”—

Explain how to solve the following IVP.

\[ -18 \, {y} + 18 \, \mathrm{u}\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 6 \]

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

Answer:

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

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

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


Example 11 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 12 πŸ”—

Explain how to solve the following IVP.

\[ -12 \, {y} + 15 \, {y'} - 36 \, \delta\left(t - 1\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 6 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{12 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 4} + \frac{6}{s^{2} - 5 \, s + 4} \]

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

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


Example 13 πŸ”—

Explain how to solve the following IVP.

\[ -24 \, {y} + 18 \, {y'} - 3 \, {y''} = 24 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8 \]

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

Answer:

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

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

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


Example 14 πŸ”—

Explain how to solve the following IVP.

\[ -27 \, {y} - 54 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3 \]

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

Answer:

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

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

\[ {y} = 2 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 3\right) \]


Example 15 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 16 πŸ”—

Explain how to solve the following IVP.

\[ 12 \, {y} + 24 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -2 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{2}{s^{2} + 4} - \frac{8 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s} \]

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

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


Example 17 πŸ”—

Explain how to solve the following IVP.

\[ -12 \, {y} = 3 \, {y''} - 48 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4 \]

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

Answer:

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

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

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


Example 18 πŸ”—

Explain how to solve the following IVP.

\[ -9 \, {y} - 3 \, {y''} + 12 \, {y'} - 18 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -2 \]

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

Answer:

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

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

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


Example 19 πŸ”—

Explain how to solve the following IVP.

\[ 18 \, {y} = -2 \, {y''} + 54 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -2 , y'(0)= 0 \]

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

Answer:

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

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

\[ {y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 2 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right) \]


Example 20 πŸ”—

Explain how to solve the following IVP.

\[ 60 \, \delta\left(t - 1\right) = -9 \, {y'} + 3 \, {y''} - 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 10 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{20 \, e^{\left(-s\right)}}{s^{2} - 3 \, s - 4} + \frac{10}{s^{2} - 3 \, s - 4} \]

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

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


Example 21 πŸ”—

Explain how to solve the following IVP.

\[ -40 \, \delta\left(t - 3\right) = -2 \, {y''} + 12 \, {y} - 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 20 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 6} + \frac{20}{s^{2} + s - 6} \]

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

\[ {y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)} \]


Example 22 πŸ”—

Explain how to solve the following IVP.

\[ 18 \, {y} + 2 \, {y''} = 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -9 \]

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

Answer:

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

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

\[ {y} = -2 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 3\right) \]


Example 23 πŸ”—

Explain how to solve the following IVP.

\[ 3 \, {y} = -3 \, {y''} - 6 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0 \]

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

Answer:

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

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

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


Example 24 πŸ”—

Explain how to solve the following IVP.

\[ 24 \, {y} - 6 \, {y'} + 72 \, \delta\left(t - 1\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 6 \]

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

Answer:

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

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

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


Example 25 πŸ”—

Explain how to solve the following IVP.

\[ 2 \, {y''} = -8 \, {y} + 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -4 \]

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

Answer:

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

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

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


Example 26 πŸ”—

Explain how to solve the following IVP.

\[ 12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2 \]

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

Answer:

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

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

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


Example 27 πŸ”—

Explain how to solve the following IVP.

\[ -12 \, {y} - 2 \, {y''} - 10 \, {y'} + 4 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2 \]

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

Answer:

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

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

\[ {y} = 2 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-2 \, t\right)} - 2 \, e^{\left(-3 \, t\right)} \]


Example 28 πŸ”—

Explain how to solve the following IVP.

\[ 24 \, \mathrm{u}\left(t - 1\right) = 3 \, {y''} + 12 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s} \]

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

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


Example 29 πŸ”—

Explain how to solve the following IVP.

\[ -12 \, {y} - 3 \, {y''} = 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -2 , y'(0)= 0 \]

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

Answer:

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

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

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


Example 30 πŸ”—

Explain how to solve the following IVP.

\[ 0 = 18 \, {y} + 2 \, {y''} - 36 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 4 , y'(0)= 0 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s} \]

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

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


Example 31 πŸ”—

Explain how to solve the following IVP.

\[ -12 \, \delta\left(t - 1\right) = 12 \, {y} - 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8 \]

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

Answer:

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

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

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


Example 32 πŸ”—

Explain how to solve the following IVP.

\[ 0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6} \]

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

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


Example 33 πŸ”—

Explain how to solve the following IVP.

\[ 2 \, {y''} + 16 \, \delta\left(t - 1\right) = 6 \, {y} + 4 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 12 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3} \]

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

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


Example 34 πŸ”—

Explain how to solve the following IVP.

\[ 0 = 8 \, {y} + 2 \, {y''} + 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 5 , y'(0)= 0 \]

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

Answer:

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

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

\[ {y} = \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 1\right) \]


Example 35 πŸ”—

Explain how to solve the following IVP.

\[ -8 \, {y} - 2 \, {y''} = 32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -2 , y'(0)= 0 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 4} - \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s} \]

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

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


Example 36 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 37 πŸ”—

Explain how to solve the following IVP.

\[ 2 \, {y''} - 32 \, {y} - 16 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 8 \]

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

Answer:

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

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

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


Example 38 πŸ”—

Explain how to solve the following IVP.

\[ -24 \, {y} - 18 \, {y'} - 24 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 4 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} + \frac{4}{s^{2} + 6 \, s + 8} \]

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

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


Example 39 πŸ”—

Explain how to solve the following IVP.

\[ 8 \, {y'} - 6 \, {y} = 2 \, {y''} - 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} - 4 \, s + 3} + \frac{6}{s^{2} - 4 \, s + 3} \]

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

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


Example 40 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 41 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 42 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 43 πŸ”—

Explain how to solve the following IVP.

\[ -3 \, {y''} - 12 \, {y} = -36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -8 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s} \]

\[ \mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{8}{s^{2} + 4} \]

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


Example 44 πŸ”—

Explain how to solve the following IVP.

\[ -18 \, {y} + 36 \, \mathrm{u}\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3 \]

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

Answer:

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

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

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


Example 45 πŸ”—

Explain how to solve the following IVP.

\[ -18 \, {y} - 2 \, {y''} = -18 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 9 \]

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

Answer:

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

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

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


Example 46 πŸ”—

Explain how to solve the following IVP.

\[ -12 \, \delta\left(t - 1\right) = -12 \, {y'} + 3 \, {y''} + 9 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8 \]

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

Answer:

\[ \mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3} \]

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

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


Example 47 πŸ”—

Explain how to solve the following IVP.

\[ -3 \, {y''} - 45 \, \delta\left(t - 1\right) = -18 \, {y} - 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -20 \]

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

Answer:

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

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

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


Example 48 πŸ”—

Explain how to solve the following IVP.

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

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

Answer:

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

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

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


Example 49 πŸ”—

Explain how to solve the following IVP.

\[ 0 = -3 \, {y''} - 18 \, {y} + 15 \, {y'} - 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4 \]

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

Answer:

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

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

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


Example 50 πŸ”—

Explain how to solve the following IVP.

\[ -18 \, {y} + 36 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 5 , y'(0)= 0 \]

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

Answer:

\[ \mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s} \]

\[ \mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{5 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} \]

\[ {y} = -2 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 5 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 3\right) \]