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All Differential Equations Resources

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Flashcards: Linear Equations

Solve the initial value problem y'' - 4y' + 8y = 0 for y(0) = 1 and y'(0) = 2

$y = 2e^{2t}\cos(2t) - 2e^{2t}\sin(2t)$

$y = 4e^{2t} - 2e^{-2t}$

$y = e^{2t}\sin(2t)$

$y = e^{2t}\cos(2t) - e^{2t}\sin(2t)$

$y = e^{2t}\cos(2t)$

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Flashcards: Linear Equations

Solve the initial value problem y'' - 4y' + 8y = 0 for y(0) = 1 and y'(0) = 2

Your answer:

$y = 2e^{2t}\cos(2t) - 2e^{2t}\sin(2t)$

Your answer:

$y = 4e^{2t} - 2e^{-2t}$

Your answer:

$y = e^{2t}\sin(2t)$

Your answer:

$y = e^{2t}\cos(2t) - e^{2t}\sin(2t)$

Correct answer:

$y = e^{2t}\cos(2t)$

Explanation:

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

$\lambda^2 - 4\lambda + 8 = 0$ We then solve the characteristic equation and find that $(\lambda - 2)^2 + 4 = 0; \lambda = 2 \pm 2i$ (Use the quadratic formula if you'd like) This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains $e^{2t}\sin(2t), e^{2t}\cos(2t)$ .

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, $y = c_1e^{2t}\sin(2t) + c_2e^{2t}\cos(2t)$ . Plugging in our initial condition, we find that 1 = c_2 . To plug in the second initial condition, we take the derivative and find that $y' = 2c_1e^{2t}(\sin(2t) + \cos(2t)) + 2e^{2t}(\cos(2t) - sin(2t))$ . Plugging in the second initial condition yields 2 = 2c_1 + 2 . Solving this simple system of linear equations shows us that

c_1 = 0; c_2 = 1

Leaving us with a final answer of $y = e^{2t}\cos(2t)$

(Note, it would have been very simple to find the right answer just by taking derivatives and plugging in, but this is not overly helpful for non-multiple choice questions)

Flashcards: Linear Equations

Find the general solution to y''' - y'' + y' - y = 0 .

$y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)$

$y = e^{t}(c_1 + c_2\sin(2t) + c_3\cos(2t))$

$y = c_1e^{t} + c_2\sin(2t) + c_3\cos(2t)$

$y = c_1e^{-t} + c_2\sin(t) + c_3\cos(t)$

$y = c_2e^{t}\sin(2t) + c_3e^{t}\cos(2t)$

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Flashcards: Linear Equations

Find the general solution to y''' - y'' + y' - y = 0 .

Your answer:

$y = e^{t}(c_1 + c_2\sin(2t) + c_3\cos(2t))$

Your answer:

$y = c_1e^{t} + c_2\sin(2t) + c_3\cos(2t)$

Your answer:

$y = c_1e^{-t} + c_2\sin(t) + c_3\cos(t)$

Your answer:

$y = c_2e^{t}\sin(2t) + c_3e^{t}\cos(2t)$

Correct answer:

$y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)$

Explanation:

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

$\lambda^3 - \lambda^2 + \lambda - 1 = 0$ To factor this, in this case we may use factoring by grouping. More generally, we may use horner's scheme/synthetic division to test possible roots. Here are both methods shown.

$\lambda^3 - \lambda^2 + \lambda - 1 = (\lambda^3 + \lambda) - (\lambda^2 + 1) = \lambda(\lambda^2 + 1) - (\lambda^2 +1) = (\lambda^2 + 1)(\lambda - 1)$

Alternatively, the rational root theorem suggests that we try -1 or 1 as a root of this equation. Using horner's scheme, we see

$1 \left |\begin{matrix}1 && -1 && 1 && -1\\ && 1 && 0 && 1 \\ \hline \end{matrix}\right.$

$1 \: \, \: \: \: \, \, \: \: \: 0 \: \: \, \, \: \: \: \, \: \:1 \: \: \, \, \: \: \: \, \: \: 0$

Which tells us the the polynomial factors into $(\lambda - 1)(\lambda^2 + 1)$ and that $\lambda = 1, \pm i$ . This means that the fundamental set of solutions is e^t, sin(t), cos(t)

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, $y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)$ . As this is not an initial value problem and just asks for the general solution, we are done.

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All Differential Equations Resources

Flashcards: Linear Equations

Flashcards: Linear Equations

Flashcards: Linear Equations

Flashcards: Linear Equations

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All Differential Equations Resources

Flashcards: Linear Equations

Flashcards: Linear Equations

Flashcards: Linear Equations

Flashcards: Linear Equations

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DMCA Complaint

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Complaint Details

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