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All Differential Equations Resources

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Flashcards: Differential Equations

Solve the following non-homogeneous differential equation. $y'' -2y' + 2y = e^{t}\sec(t)$

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + te^t\sin(t) + e^t\cos(t)\ln|\cos(t)|$

$y = c_1e^t\sin(t) + c_2e^t\cos(t) - e^t\cos(t)\ln|\sec(t)|$

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + 4te^t\sin(t)$

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + e^t\sin(t) + e^t\cos(t)\tan(t)$

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + 4te^t\cos(t)$

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Flashcards: Differential Equations

Solve the following non-homogeneous differential equation. $y'' -2y' + 2y = e^{t}\sec(t)$

Your answer:

$y = c_1e^t\sin(t) + c_2e^t\cos(t) - e^t\cos(t)\ln|\sec(t)|$

Your answer:

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + 4te^t\sin(t)$

Your answer:

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + e^t\sin(t) + e^t\cos(t)\tan(t)$

Your answer:

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + 4te^t\cos(t)$

Correct answer:

$y = c_1e^t\sin(t) + c_2e^t\cos(t) + te^t\sin(t) + e^t\cos(t)\ln|\cos(t)|$

Explanation:

Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of y'' -2y' + 2y = 0 is $\lambda^2 - 2\lambda + 2 = 0$ , with solutions of $\frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$ . This means that $y_1 = e^t\sin(t)$ and $y_2 = e^{t}\cos(t)$ .

To do variation of parameters, we will need the Wronskian, $W(y_1, y_2) = \begin{vmatrix} y_1 & y_2\\ y_1'&y_2' \end{vmatrix} = e^t\sin(t)(e^t\cos(t) - e^t\sin(t)) - e^t\cos(t)(e^t\sin(t) + e^t\cos(t)) = -e^{2t}(\sin(t)^2 + \cos(t)^2) = -e^{2t}$

Variation of parameters tells us that the coefficient in front of y_i is $\int\frac{f(t)W_i(t)}{W(t)}dt$ where W_i is the Wronskian with the i'th row replaced with all 0's and a 1 at the bottom. In the 2x2 case this means that

$y_p = -y_1\int \frac{f(t)y_2}{W(t)}dt + y_2\int \frac{f(t)y_1}{W(t)}dt$ . Plugging in, the first half simplifies to

$-e^t\sin(t)\int \frac{(e^t\sec(t))(e^t\cos(t))}{-e^{2t}}dt = e^t\sin(t)\int 1dt = te^t\sin(t)$

and the second half becomes $e^t\cos(t)\int \frac{(e^t\sec(t))(e^t\sin(t))}{-e^{2t}}dt = -e^t\cos(t)\int \tan(t)dt = e^t\cos(t)\ln|\cos(t)|$

Putting these together with the complementary solution, we have a general solution of $y = c_1e^t\sin(t) + c_2e^t\cos(t) + te^t\sin(t) + e^t\cos(t)\ln|\cos(t)|$

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All Differential Equations Resources

Flashcards: Differential Equations

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Flashcards: Differential Equations

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Flashcards: Differential Equations

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