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All Differential Equations Resources

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Flashcards: Differential Equations

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

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

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

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

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

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

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Flashcards: Differential Equations

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

Your answer:

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

Your answer:

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

Your answer:

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

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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Flashcards: Differential Equations

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Flashcards: Differential Equations

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