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Differential Equations : Differential Equations

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Example Questions

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Example Question #4 : Matrix Exponentials

Use the definition of matrix exponential,

$e^{At}=I+At+A^2\frac{t^2}{2!}+...+A^k\frac{t^k}{k!}+...=\sum_{k=0}^\infty A^k\frac{t^k}{k!}$

to compute $e^{At}$ of the following matrix.

$A=\begin{pmatrix} 0& 1\\ 2&1 \end{pmatrix}$

Possible Answers:

$e^{At}=\begin{pmatrix} 0&e^t \\ e^{2t}&e^t \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&e^t \\ e^{2t}&2e^t \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&1\\ e^{2t}&e^t \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&e^t \\ 2e^{t}&e^t \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&e^t \\ e^{2t}&1\end{pmatrix}$

Correct answer:

$e^{At}=\begin{pmatrix} 0&e^t \\ e^{2t}&e^t \end{pmatrix}$

Explanation:

Given the matrix,

$A=\begin{pmatrix} 0& 1\\ 2&1 \end{pmatrix}$

and using the definition of matrix exponential,

$e^{At}=I+At+A^2\frac{t^2}{2!}+...+A^k\frac{t^k}{k!}+...=\sum_{k=0}^\infty A^k\frac{t^k}{k!}$

calculate

$A^2=AA=\begin{pmatrix} 0&1 \\ 2&1 \end{pmatrix} \begin{pmatrix} 0&1 \\ 2&1 \end{pmatrix}=\begin{pmatrix} 0& 1\\ 4& 1 \end{pmatrix}$

$A^3=A^2A=\begin{pmatrix} 0&1 \\ 4&1 \end{pmatrix} \begin{pmatrix} 0&1 \\ 2&1 \end{pmatrix}=\begin{pmatrix} 0&1 \\ 8&1 \end{pmatrix}$

Therefore

$A^k=\begin{pmatrix} 0&1 \\ 2^k&1 \end{pmatrix}; k=1,2,3,...$

$e^{At}=I+\frac{A}{1!}t+\frac{A^2}{2!}t^2+...$

$\begin{pmatrix} 0& 1\\ 2&1 \end{pmatrix}+\frac{1}{1!}\begin{pmatrix} 0& 1\\ 4&1 \end{pmatrix}+\frac{1}{3!}\begin{pmatrix} 0& 1\\ 8&1 \end{pmatrix}$

$e^{At}=\begin{pmatrix} 0&e^t \\ e^{2t}&e^t \end{pmatrix}$

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Example Question #1 : Matrix Exponentials

Calculate the matrix exponential, $e^{tA}$ , for the following matrix: $A = \begin{bmatrix} -7& 3\\-1 &-3 \end{bmatrix}$ .

Possible Answers:

$\frac{1}{2}\begin{bmatrix} 3e^{-6t}-e^{-4t} & -3e^{-6t} + 3 e^{-4t} \\ e^{-6t}-e^{-4t}&-e^{-6t} + 3e^{-4t} \end{bmatrix}$

$\begin{bmatrix} e^{-6t}-3e^{-4t} & -5e^{-6t} + e^{-4t} \\ e^{-6t}-e^{-4t}&-e^{-6t} + 8e^{-4t} \end{bmatrix}$

$\frac{1}{2}\begin{bmatrix} 3e^{-5t}-e^{-3t} & -3e^{-5t} + 3 e^{-3t} \\ e^{-5t}-e^{-3t}&-e^{-5t} + 3e^{-3t} \end{bmatrix}$

$\begin{bmatrix} 10e^{-5t}-2e^{-3t} & -e^{-5t} + 3 e^{-3t} \\ 2e^{-5t}-e^{-3t}&e^{-5t} + 4e^{-3t} \end{bmatrix}$

$\begin{bmatrix} 3e^{-5t}-e^{-3t} & -3e^{-5t} + 3 e^{-3t} \\ e^{-5t}-e^{-3t}&-e^{-5t} + 3e^{-3t} \end{bmatrix}$

Correct answer:

$\frac{1}{2}\begin{bmatrix} 3e^{-6t}-e^{-4t} & -3e^{-6t} + 3 e^{-4t} \\ e^{-6t}-e^{-4t}&-e^{-6t} + 3e^{-4t} \end{bmatrix}$

Explanation:

To get the matrix exponential, we will have to diagonalize the matrix, which requires us to find the eigenvalues and eigenvectors. Thus, we have

$\begin{vmatrix} \lambda+7 &-3 \\ 1& \lambda +3 \end{vmatrix} = (\lambda+7)(\lambda+3) +3 = \lambda^2 + 10\lambda +24 = (\lambda + 6)(\lambda + 4) = 0$

Using $\lambda = -4, -6$ , we then find the eigenvectors by solving for the eigenspace.

$\lambda = -4$

$\begin{bmatrix} 3 & -3 &\vline &0 \\ 1& -1 &\vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 &\vline &0 \\ 0& 0 &\vline & 0 \end{bmatrix}$

This has solutions x_1 - x_2 = 0 , or $x_1\begin{bmatrix} 1\\ 1 \end{bmatrix}$ . So a suitable eigenvector is simply $\begin{bmatrix} 1\\1 \end{bmatrix}$ .

Repeating for $\lambda = -6$ ,

$\begin{bmatrix} 1 & -3 &\vline &0 \\ 1& -3 &\vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -3 &\vline &0 \\ 0& 0 &\vline & 0 \end{bmatrix}$

This has solutions x_1 - 3x_2 = 0 , and thus a suitable eigenvector is $\begin{bmatrix} 3\\1 \end{bmatrix}$ .

Thus, we have $P = \begin{bmatrix} 1 & 3\\ 1 & 1 \end{bmatrix}$ , $D = \begin{bmatrix} -4 &0 \\ 0& -6 \end{bmatrix}$ , and using the inverse formula for 2x2 matrices, $P^{-1} = -\frac{1}{2}\begin{bmatrix} 1 &-3 \\ -1 &1 \end{bmatrix}$ . Now we just take the matrix exponential of and multiply the three matrices back together. Thus,

$e^{tA} = Pe^{tD}P^{-1} = -\frac{1}{2}\begin{bmatrix} 1 &3 \\ 1& 1 \end{bmatrix}\begin{bmatrix} e^{-4t} &0 \\ 0& e^{-6t} \end{bmatrix}\begin{bmatrix} 1 &-3 \\ -1&1 \end{bmatrix}$

Multiplying these out yields $e^{tA} = \frac{1}{2}\begin{bmatrix} 3e^{-6t}-e^{-4t} & -3e^{-6t} + 3 e^{-4t} \\ e^{-6t}-e^{-4t}&-e^{-6t} + 3e^{-4t} \end{bmatrix}$

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Example Question #1 : Matrix Exponentials

Find the general solution to

$y^{'''} + y^{''} + 3y' - 5y = 0$

Possible Answers:

$y(x) = C_1e^x + C_2e^{-x}\cos{2x}$

$y(x) = C_1e^x + C_2e^{-x}\sin{2x}$

$y(x) = C_1e^x + C_2e^{-x}\cos{2x} + C_3e^{-x}\sin{2x}$

$y(x) = C_1e^x + C_2e^{-x}\cos{x} + C_3e^{-x}\sin{x}$

None of the other answers

Correct answer:

$y(x) = C_1e^x + C_2e^{-x}\cos{2x} + C_3e^{-x}\sin{2x}$

Explanation:

The auxiliary equation is

r^3+r^2+3r-5 = (r-1)(r^2+2r+5) = 0

The roots are

$r = 1, -1 \pm 2i$

Our solution is

$y(x) = C_1e^x + C_2e^{-x}\cos{2x} + C_3e^{-x}\sin{2x}$

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Example Question #1 : The Laplace Transform

Find the Laplace Transform for the following function.

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}$

Possible Answers:

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+3}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s<-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s-5}, s>-5$

Correct answer:

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>-5$

Explanation:

To find the Laplace Transform for the following function

$\mathcal{L}\begin{Bmatrix} e^{-5t}\end{Bmatrix}$

use the transforms for basic functions which state,

$\mathcal{L}\begin{Bmatrix} 1 \end{Bmatrix}=\frac{1}{s}$

$\mathcal{L}\begin{Bmatrix} e^{at} \end{Bmatrix}=\frac{1}{s-a}$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\int_0^\infty e^{-st}e^{-5t}dt=\int_0^\infty e^{-(s+5)t} dt$

$\\=\frac{-e^{-(s+5)t}}{s+5}\bigg|_0^\infty \\\\=\frac{1}{s+5}, s>-5$

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Example Question #2 : The Laplace Transform

Find the Laplace transform of the periodic function.

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\\ 0, & 2\leq4 \end{matrix}\right.$

Possible Answers:

$\frac{1}{s(1+e^{-2s})(1-e^{-s})}$

$\frac{1}{s(1+e^{2s})(1+e^{-s})}$

$\frac{1}{s(1-e^{-2s})(1+e^{-s})}$

$\frac{1}{s(1+e^{-2s})(1+e^{s})}$

$\frac{1}{s(1+e^{-2s})(1+e^{-s})}$

Correct answer:

$\frac{1}{s(1+e^{-2s})(1+e^{-s})}$

Explanation:

This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

T=4 .

and furthermore,

f(t+4)=f(t)

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) \end{Bmatrix}=\frac{1}{1-e^{-sT}}\int_0^T e^{-sT}f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) \end{Bmatrix}=\frac{1}{1-e^{-4s}}\int_0^4 e^{-st}S(t)dt$

$\\=\frac{1}{1-e^{-4s}}\left[ \int_0^2 e^{-st}\cdot 1 dt+\int_2^4 e^{-st}\cdot 0dt \right] \\\\=\frac{1}{1-e^{-4s}}\frac{1-e^{-s}}{s}\\\\\leftarrow 1-e^{-4s}=(1+e^{-2s})(1-e^{-2s})\\\\\leftarrow (1-e^{-2s})=(1+e^{-s})(1-e^{-s}) \\\\=\frac{1-e^{-s}}{s(1+e^{-2s})(1-e^{-s})(1+e^{-s})} \\\\=\frac{1}{s(1+e^{-2s})(1+e^{-s})}$

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Differential Equations : Differential Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #4 : Matrix Exponentials

Example Question #1 : Matrix Exponentials

Example Question #1 : Matrix Exponentials

Example Question #1 : The Laplace Transform

Example Question #2 : The Laplace Transform

All Differential Equations Resources

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