Differential Equations - Second-Order Boundary-Value Problems

Find the solutions to the second order boundary-value problem. , , $y\left(\frac{\pi}{2}\right) = 1$ .

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

$y = 5e^t\sin(t) + 3e^t\cos(t)$

There are no solutions to the boundary value problem.

$y = 10e^{2t}\sin(2t)$

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

Explanation

The characteristic equation of is $\lambda^2 - 2\lambda + 2 = 0$ , with solutions of $\frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$ . Thus, the general solution to the homogeneous problem is $y = c_1e^t\sin(t) + c_2e^t\cos(t)$ . Plugging in our conditions, we find that , so that $y = c_1e^t\sin(t)$ . Plugging in our second condition, we find that $y\left(\frac{\pi}{2}\right) = c_1e^{\frac{\pi}{2}} = 1$ and that $c_1 = e^{-\frac{\pi}{2}}$ .

Thus, the final solution is $y = e^{t -\frac{\pi}{2}}\sin(t)$ .

Find the solutions to the second order boundary-value problem. , , $y\left(2\pi\right) = 1$ .

There are no solutions to the boundary value problem.

$y = 3\sin(3t)$

$y = \sin(3t) + 2 \cos(3t)$

$y = 4\sin(3t) - \cos(3t)$

$y = 3\cos(3t)$

Explanation

The characteristic equation of is $\lambda^2 + 9 = 0$ with solutions of $\pm 3i$ . This tells us that the solution to the homogeneous equation is $y = c_1\sin(3t) + c_2\cos(3t)$ . Plugging in our conditions, we find that so that $y = c_1\sin(3t)$ . Plugging in our second condition, we have $y(2\pi) = 0 = 1$ which is obviously false.

This problem demonstrates the important distinction between initial value problems and boundary value problems: Boundary value problems don't always have solutions. This is one such case, as we can't find that satisfy our conditions.

Second-Order Boundary-Value Problems

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