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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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