Differential Equations Recap (Polytechnic)

First-Order Differential Equations

Problem 1. Solve the separable differential equation $\displaystyle \frac{\mathrm dy}{\mathrm dx} = x^2 y$ .

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Solution. By the method of separable variables,

$\begin{aligned} \int \frac 1y\, \mathrm dy &= \int x^2\, \mathrm dx,\\ \ln |y| &= \frac{x^3}{3} + C \\ |y| &= e^{x^3/3 + C} \\ y &= \pm e^C \cdot e^{x^3 / 3} \\ y &= Ae^{x^3/3},\end{aligned}$

where we consolidated the arbitrary constant $A = \pm e^C$ .

Problem 2. Solve the linear differential equation $\displaystyle \frac{\mathrm dy}{\mathrm dx} + 2y = e^{3x}$ .

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Solution. By the method of integrating factors, we identify $P(x) = 2$ and $Q(x) = e^{3x}$ . The integrating factor $I(x)$ is computed via

$I(x) = e^{\int P(x)\, \mathrm dx} = e^{\int 3\, \mathrm dx} = e^{3x}.$

(In our notes, we denoted $\mu(x) \equiv I(x)$ . This solution uses the notation that is used in the school that I teach.) The general solution is then given by

$\begin{aligned} y I(x) &= \int I(x)Q(x)\, \mathrm dx \\ y \cdot e^{2x} &= \int e^{3x} \cdot e^{2x}\, \mathrm dx\\ &= \int e^{5x}\, \mathrm dx\\ &= \frac 15 e^{5x} + C \end{aligned}$

Second-Order Differential Equations

Problem 3. Solve the differential equation $y'' - 5y' + 6y = 0$ .

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Solution. The characteristic equation

$m^2 - 5m + 6 = 0$

has solutions $m = 2, 3$ , which are real and distinct. Therefore, the general solution is given by

$y = C_1 e^{2x} + C_2 e^{3x}.$

Problem 4. Evaluate $\displaystyle \frac 1{\mathcal D^2 + 1} ( 2 + e^{2x} + \cos(x) + \sin(2x) )$ .

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Solution. By the linearity of inverse- $\mathcal D$ operators,

$\begin{aligned} &\frac 1{\mathcal D^2 + 1}\left( + e^{2x} + \cos(x) + \sin(2x) \right) \\ &= \frac 1{\mathcal D^2 + 1}(2) + \frac 1{\mathcal D^2 + 1} (e^{2x}) + \frac 1{\mathcal D^2 + 1}(\cos(x)) + \frac 1{\mathcal D^2 + 1}(\sin(2x)) \\ &= \frac 1{0^2 + 1} \cdot 2 + \frac 1{2^2 + 1} \cdot e^{2x} + \frac{x \sin(x)}{2 \cdot 1} + \frac 1{-2^2 + 1} \cdot \sin(2x) \\ &= 2 + \frac 1{5} e^{2x} + \frac{1}{2}x \sin(x) - \frac 1{3} \sin(2x). \end{aligned}$

Problem 5. Evaluate $\displaystyle \frac 1{\mathcal D^2 - 4\mathcal D - 5}(e^{2x} \sin(x))$ .

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Solution. Applying the shift property for inverse- $\mathcal D$ operators,

$\begin{aligned}\frac 1{\mathcal D^2 - 4\mathcal D - 5}(e^{2x} \sin(x)) &= e^{2x} \cdot \frac 1{(\mathcal D + 2)^2 - 4(\mathcal D + 2) - 5}(\sin(x)) \\ &= e^{2x} \cdot \frac 1{(\mathcal D^2 + 4 \mathcal D + 4) - (4\mathcal D + 8) - 5}(\sin(x)) \\ &= e^{2x} \cdot \frac 1{\mathcal D^2 - 9}(\sin(x)) \\ &= e^{2x} \cdot \frac 1{-1^2 - 9} \cdot \sin(x) \\ &= -\frac 1{10} e^{2x} \sin(x). \end{aligned}$

Problem 6. Solve the differential equation $\displaystyle \frac{\mathrm d^2 y}{\mathrm d x^2} - 4 \frac{\mathrm d y}{\mathrm d x} - 5y = 10 e^{2x} \sin(x)$ .

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Solution. To obtain the complementary function $y_{\mathrm C}$ , we solve the equation

$\displaystyle \frac{\mathrm d^2 y}{\mathrm d x^2} - 4 \frac{\mathrm d y}{\mathrm d x} - 5y = 0.$

The characteristic equation

$m^2 - 4m - 5 = 0$

has solutions $m = -1, 5$ , which are real and distinct. Therefore, the general solution is given by

$y = C_1 e^{-x} + C_2 e^{5x}.$

For the particular integral, we re-write the problem using $\mathcal D$ -notation

$(\mathcal D^2 - 4\mathcal D - 5)y_{\mathrm P} = 10 e^{2x} \sin(x)$

and use Problem 5 to obtain

$\begin{aligned} y_{\mathrm P} &= \frac 1{\mathcal D^2 - 4\mathcal D - 5} (10 e^{2x} \sin(x)) \\ &= 10 \cdot \frac 1{\mathcal D^2 - 4\mathcal D - 5} (e^{2x} \sin(x)) \\ &= 10 \cdot -\frac 1{10} e^{2x} \sin(x) \\ &= -e^{2x} \sin(x). \end{aligned}$

Consolidating, the general solution is given by

$y = y_{\mathrm P} + y_{\mathrm C} = -e^{2x} \sin(x) + C_1 e^{-x} + C_2 e^{5x}.$

Laplace Transforms

Problem 7. Evaluate $\mathcal L\{3 + 2e^{3t} + t^3 + \sin (2t)\}$ .

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Solution. By the linearity of taking Laplace transforms,

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

Problem 8. Evaluate $\displaystyle \mathcal L^{-1}\left\{ \frac 2{s^2} + \frac 2{s-1} + \frac 4{s^2+4} + \frac{5s}{s^2 + 16} \right\}$ .

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Solution. By the linearity of taking inverse Laplace transforms,

$\begin{aligned} & \mathcal L^{-1}\left\{ \frac 2{s^2} + \frac 2{s-1} + \frac 4{s^2+4} + \frac{5s}{s^2 + 16} \right\} \\ &= 2 \cdot \mathcal L^{-1}\left\{ \frac {1!}{s^{1+1}} \right\} + 2 \cdot \mathcal L^{-1}\left\{ \frac 1{s-1} \right\} \\ &\phantom{==} + 2 \cdot \mathcal L^{-1}\left\{ \frac 2{s^2+2^2} \right\} + 5 \cdot \mathcal L^{-1}\left\{ \frac s{s^2+4^2} \right\} \\ &= 2 t + 2 e^t + 2 \sin(2t) + 5 \cos(4t). \end{aligned}$

Problem 9. The function $f$ is defined by

$f(t) = \begin{cases} 0, & t < 2, \\ 3, & 2 \leq t < 5, \\ 4, & 5 \leq t < 7, \\ 2, & t \geq 7.\end{cases}$

Evaluate $\mathcal L\{e^{2t}f(t)\}$ .

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Solution. By the shift theorems,

$\mathcal L\{e^{2t} f(t)\} = F(s - 2),$

where $\mathcal L\{f(t)\} = F(s)$ . To evaluate $f(t)$ , we use the jump technique to obtain

$\begin{aligned} f(t) &= 0 + (3-0) \cdot U(t-2) + (4-3) \cdot U(t-5) + (2-4) \cdot U(t-7) \\ &= 3 U(t-2) + U(t-5) - 2 U(t-7). \end{aligned}$

By the linearity of taking Laplace transforms,

$\begin{aligned} F(s) = \mathcal L \{f(t) \} &= \mathcal L \{ 3 U(t-2) + U(t-5) - 2 U(t-7) \} \\ &= 3 \cdot \mathcal L \{ U(t-2) \} + \mathcal L \{ U(t-5) \} - 2 \cdot \mathcal L \{U(t-7) \} \\ &= 3 \cdot \frac{e^{-2s}}{s} + \frac{e^{-5s}}{s} - 2 \cdot \frac{e^{-7s}}{s} \\ &= \frac{3e^{-2s} + e^{-5s} - 2e^{-7s}}{s}. \end{aligned}$

Therefore,

$\displaystyle \mathcal L\{e^{2t} f(t)\} = F(s - 2) = \frac{3e^{-2(s-2)} + e^{-5(s-2)} - 2e^{-7(s-2)}}{s-2}.$

Problem 10. Solve the initial value problem

$y'' + y = \delta(t-2),\quad y(0)=y'(0) = 0.$

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Solution. Taking Laplace transforms on both sides and applying linearity,

$\begin{aligned} \mathcal L\{ y'' \} + \mathcal L \{ y \} &= \mathcal L \{ \delta(t-2) \} \end{aligned}$

To evaluate each expression, $\mathcal L\{y\} = Y(s)$ , $\mathcal L \{ \delta(t-2) \} = e^{-2s}$ , and

$\mathcal L\{ y'' \} = s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s).$

Hence,

$\begin{aligned} \mathcal L\{ y'' \} + \mathcal L \{ y \} &= \mathcal L \{ \delta(t-2) \} \\ s^2 Y(s) + Y(s) &= e^{-2s} \\ (s^2 + 1) Y(s) &= e^{-2s} \\ Y(s) &= e^{-2s} \cdot \frac{1}{s^2+1}. \end{aligned}$