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All Differential Equations Resources

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Want to review Differential Equations but don’t feel like sitting for a whole test at the moment? Varsity Tutors has you covered with thousands of different Differential Equations flashcards! Our Differential Equations flashcards allow you to practice with as few or as many questions as you like. Get some studying in now with our numerous Differential Equations flashcards.

Flashcards: Linear Equations

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

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

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

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

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

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

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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:

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

Your answer:

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

Your answer:

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

Your answer:

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

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 .

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

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

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

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

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

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Flashcards: Linear Equations

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

Your answer:

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

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:

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

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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