KindiakMath

Mid-Term Rally

Big Idea

Laplace transforms are linear transformations:

\begin{aligned} \mathcal L\{f(t) + g(t)\} &= \mathcal L\{f(t)\} + \mathcal L\{g(t)\}, \\ \mathcal L\{ c \cdot f(t)\}& = c \cdot \mathcal L\{f(t)\}. \end{aligned}

Often, we denote F(s) = \mathcal L\{f(t)\}, so that inverse Laplace transforms are linear transformations too:

\begin{aligned} \mathcal L^{-1} \{F(s) + G(s)\} &= \mathcal L^{-1} \{F(s)\} + \mathcal L^{-1} \{G(s)\} \\ \mathcal L^{-1}\{ c \cdot F(s)\} &= c \cdot \mathcal L^{-1}\{F(s)\}. \end{aligned}

We will also revise previously-discussed topics.

Questions

Question 1. Evaluate \mathcal L \{ 2 - e^{2t} + \sin(5t) - t^2\}.

(Click for Solution)

Solution. By the linearity of Laplace transforms,

\begin{aligned} \mathcal L \{ 2 - e^{2t} + \sin(5t) - t^2 \} &= 2 \cdot \mathcal L\{ 1 \} - \mathcal L\{ e^{2t} \} + \mathcal L\{ \sin(5t) \} - \mathcal L\{ s^2 \} \\ &= 2 \cdot \frac 1s - \frac 1{s-2} + \frac{5}{s^2+5^2} - \frac{2!}{s^{2+1}} \\ &= \frac 2s - \frac 1{s-2} + \frac 5{s^2 + 25} - \frac 2{s^3}.\end{aligned}

Question 2. Evaluate

\displaystyle \mathcal L^{-1} \left\{ \frac 1{s-3} + \frac{s}{s^2 + 4} + \frac 6{s^4} + \frac 1s \right\}.

(Click for Solution)

Solution. By the linearity of inverse Laplace transforms,

\begin{aligned} &\mathcal L^{-1} \left\{ \frac 1{s-3} + \frac{s}{s^2 + 4} + \frac 6{s^4} + \frac 1s \right\} \\ &= \mathcal L^{-1} \left\{ \frac 1{s-3} \right\} + \mathcal L^{-1} \left\{ \frac s{s^2+2^2} \right\} + \mathcal L^{-1} \left\{ \frac{3!}{s^{3+1}} \right\} + \mathcal L^{-1} \left\{ \frac 1s \right\} \\ &= e^{3t} + \cos(2t) + t^3 + 1. \end{aligned}

Question 3. Use Laplace transforms to evaluate t^2 * \cos(2t).

Hint. Recall that \mathcal L\{f * g\} = \mathcal L\{f\} \cdot \mathcal L\{g\}.

(Click for Solution)

Solution. Taking the Laplace transform of a convolution,

\begin{aligned} \mathcal L\{ t^2 * \cos(2t) \} &= \mathcal L\{t^2 \} \cdot \mathcal L\{\cos(2t) \} \\ &= \frac{2!}{s^{2+1}} \cdot \frac{s}{s^2 + 2^2} \\ &= \frac 2{s^3} \cdot \frac{s}{s^2 + 4} \\ &= \frac 2{s^2(s^2 + 4)}. \end{aligned}

Therefore,

\begin{aligned} t^2 * \cos(2t) &= \mathcal L^{-1} \left\{ \frac 2{s^2(s^2 + 4)} \right\}. \end{aligned}

Performing partial fraction decomposition (see Problem 15 in this post for the general technique),

\begin{aligned} \frac{2}{s^2(s^2+4)} &= \frac 12 \left( \frac{1}{s^2} - \frac{1}{s^2+4} \right) = \frac 12 \cdot \frac {1}{s^2} - \frac 14 \cdot \frac{2}{s^2+4}. \end{aligned}

By the linearity of inverse Laplace transforms,

\begin{aligned} t^2 * \cos(2t) &= \mathcal L^{-1} \left\{ \frac 2{s^2(s^2 + 4)} \right\} \\ &= \mathcal L^{-1} \left\{ \frac 12 \cdot \frac 1{s^2} - \frac 14 \cdot \frac{2}{s^2+4} \right\} \\ &= \frac 12 \cdot \mathcal L\left\{ \frac 1{s^2} \right\} - \frac 14 \cdot \mathcal L^{-1} \left\{ \frac{2}{s^2+4} \right\} \\ &= \frac 12 \cdot \mathcal L\left\{ \frac {1!}{s^{1+1}} \right\} - \frac 14 \cdot \mathcal L^{-1} \left\{ \frac{2}{s^2+2^2} \right\} \\ &= \frac 12 t - \frac 14 \sin(2t). \end{aligned}

Remark. Without Laplace transforms, you will need to integrate by parts to obtain

\begin{aligned} \int u \cos(2u)\, \mathrm du &= \frac 12 u \sin(2u) + \frac 14 \cos(4u) + C, \\ \int u^2 \cos(2u)\, \mathrm du &= \frac 12 u^2 \sin(2u) + \frac 12 u \cos(2u) - \frac 14 \sin(2u) + C ,\end{aligned}

then use the definition of convolution to get

\begin{aligned} t^2 * \cos(2t) &= \int_0^t (t-u)^2 \cos(2u)\, \mathrm du \\ &= t^2 \int_0^t \cos(2u)\, \mathrm du - 2u \int_0^t u\cos(2u)\, \mathrm du + \int_0^t u^2 \cos(2u)\, \mathrm du \\ &= t^2 \cdot \frac 12 \sin(2t) - 2t \cdot \left( \frac 12 t \sin(2t) + \frac 14 \cos(2t) - \frac 14 \right) \\ &\phantom{==} + \left( \frac 12 t^2 \sin(2t) + \frac 12 t \cos(2t) - \frac 14 \sin(2t) \right) \\ &= \frac 12 t - \frac 14 \sin(2t). \end{aligned}

Question 4. Solve the differential equation

\displaystyle y'' + 2y' - 3y = 4e^x + 5e^{2x}.

(Click for Solution)

Solution. Rewriting in \mathcal D-notation,

(\mathcal D^2 + 2 \mathcal D - 3)y = 4e^x + 5e^{2x}.

The general solution is given by y = y_{\mathrm P} + y_{\mathrm C}. To obtain the complementary function y_{\mathrm C}, we solve the equation

(\mathcal D^2 + 2 \mathcal D - 3)y = 0.

Since the characteristic equation m^2 + 2m - 3 = 0 has real and distinct auxiliary roots m = -3, 1, the general solution is given by

y_{\mathrm C} = C_1 e^{-3x} + C_2 e^x.

To evaluate y_{\mathrm P}, we first use linearity to compute

\begin{aligned} y_{\mathrm P} &= \frac{1}{\mathcal D^2 + 2 \mathcal D - 3}(4e^x + 5e^{2x}) \\ &= 4 \cdot \underbrace{\frac{1}{\mathcal D^2 + 2 \mathcal D - 3}(e^x)}_{y_1} +\, 5 \cdot \underbrace{\frac{1}{\mathcal D^2 +2 \mathcal D - 3}(e^{2x})}_{y_2}. \end{aligned}

To shorten presentation, the general solution is y = 4y_1 + 5y_2 + y_{\mathrm C}. For the simpler term y_2,

\displaystyle y_2 = \frac{1}{\mathcal D^2 + 2 \mathcal D - 3}(e^{2x}) = \frac 1{2^2 + 2 \cdot 2 - 3} \cdot e^{2x} = \frac{1}{5} e^{2x}.

For the challenging term y_1, we first replace each \mathcal D with \mathcal D + 1 to obtain

\begin{aligned} (\mathcal D + 1)^2 + 2 (\mathcal D + 1) - 3 &= (\mathcal D^2 + 2\mathcal D + 1) + (2\mathcal D + 2) - 3\\ &= \mathcal D^2 + 4 \mathcal D = \mathcal D(\mathcal D+4). \end{aligned}

Therefore,

\begin{aligned} y_1 = \frac{1}{\mathcal D^2 + 2 \mathcal D - 3}(e^{x}) &= e^{x} \cdot \frac{1}{\mathcal D(\mathcal D+4)} (1) \\ &= e^{x} \cdot \frac 1{\mathcal D}\left(\frac 1{\mathcal D+4} (1)\right) \\ &= e^{x} \cdot \frac 1{\mathcal D}\left( \frac 1{0+4} \cdot 1 \right) \\ &= \frac 14 e^x \cdot \frac 1{\mathcal D}(1) = \frac 14 e^x \cdot x. \end{aligned}

Consolidating all results,

y = 4y_1 + 5y_2 + y_{\mathrm C} = x e^x + e^{2x} + (C_1 e^{-3x} + C_2 e^x).

—Joel Kindiak, 18 Apr 25, 1536H

,

Published by

joelkindiak

Leave a comment