Second-Order Boundary-Value Problems - Differential Equations

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Question

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

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Answer

The characteristic equation of is , 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)$ .

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Question

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

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Answer

The characteristic equation of is 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.

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Question

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

Tap to reveal answer

Answer

The characteristic equation of is , 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)$ .

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Question

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

Tap to reveal answer

Answer

The characteristic equation of is 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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

The characteristic equation of is , 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)$ .

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Question

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

Tap to reveal answer

Answer

The characteristic equation of is 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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

The characteristic equation of is , 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)$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

The characteristic equation of is 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.

← Didn't Know|Knew It →

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