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Flashcards: Higher-Order Differential Equations

For the initial value problem

$x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$ $\displaystyle x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$

$\displaystyle y(x_0) = 1, y'(x_0) = 0, y''(x_0) = 7$

Which if the following intervals containing x_0 $\displaystyle x_0$ do NOT guarantees the existence of a unique solution?

(-5,0) $\displaystyle (-5,0)$

(1,5) $\displaystyle (1,5)$

(0,1) $\displaystyle (0,1)$

(-4,-3) $\displaystyle (-4,-3)$

(0,2) $\displaystyle (0,2)$

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Flashcards: Higher-Order Differential Equations

For the initial value problem

$x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$ $\displaystyle x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$

$\displaystyle y(x_0) = 1, y'(x_0) = 0, y''(x_0) = 7$

Which if the following intervals containing x_0 $\displaystyle x_0$ do NOT guarantees the existence of a unique solution?

(-5,0) $\displaystyle (-5,0)$

(1,5) $\displaystyle (1,5)$

(0,1) $\displaystyle (0,1)$

(-4,-3) $\displaystyle (-4,-3)$

(0,2) $\displaystyle (0,2)$

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Flashcards: Higher-Order Differential Equations

For the initial value problem

$x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$ $\displaystyle x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$

$\displaystyle y(x_0) = 1, y'(x_0) = 0, y''(x_0) = 7$

Which if the following intervals containing x_0 $\displaystyle x_0$ do NOT guarantees the existence of a unique solution?

Your answer:

(-5,0) $\displaystyle (-5,0)$

Your answer:

(1,5) $\displaystyle (1,5)$

Your answer:

(0,1) $\displaystyle (0,1)$

Your answer:

(-4,-3) $\displaystyle (-4,-3)$

Correct answer:

(0,2) $\displaystyle (0,2)$

Explanation:

Putting the equation in standard form we get that

$y^{'''} - \frac{3}{(x-1)}y^{''} + \frac{6x}{(x-1)}y' - \frac{\cos{x}}{x(x-1)}y = \frac{\sqrt{x+5}}{x(x-1)}$ $\displaystyle y^{'''} - \frac{3}{(x-1)}y^{''} + \frac{6x}{(x-1)}y' - \frac{\cos{x}}{x(x-1)}y = \frac{\sqrt{x+5}}{x(x-1)}$

We need to find where these coefficients are simultaneously continuous. This is where

$(-5,0), (0,1), (1,\infty)$ $\displaystyle (-5,0), (0,1), (1,\infty)$. The choice that is not a subset of these is (0,2) $\displaystyle (0,2)$

Flashcards: Higher-Order Differential Equations

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

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

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

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

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

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

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Flashcards: Higher-Order Differential Equations

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

Your answer:

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

Your answer:

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

Your answer:

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

Your answer:

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

Correct answer:

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

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:

$\lambda^2 - 4\lambda + 8 = 0$ $\displaystyle \lambda^2 - 4\lambda + 8 = 0$ We then solve the characteristic equation and find that $(\lambda - 2)^2 + 4 = 0; \lambda = 2 \pm 2i$ $\displaystyle (\lambda - 2)^2 + 4 = 0; \lambda = 2 \pm 2i$ (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 $e^{2t}\sin(2t), e^{2t}\cos(2t)$ $\displaystyle e^{2t}\sin(2t), e^{2t}\cos(2t)$.

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, $y = c_1e^{2t}\sin(2t) + c_2e^{2t}\cos(2t)$ $\displaystyle y = c_1e^{2t}\sin(2t) + c_2e^{2t}\cos(2t)$. Plugging in our initial condition, we find that 1 = c_2 $\displaystyle 1 = c_2$. To plug in the second initial condition, we take the derivative and find that $y' = 2c_1e^{2t}(\sin(2t) + \cos(2t)) + 2e^{2t}(\cos(2t) - sin(2t))$ $\displaystyle y' = 2c_1e^{2t}(\sin(2t) + \cos(2t)) + 2e^{2t}(\cos(2t) - sin(2t))$. Plugging in the second initial condition yields $\displaystyle 2 = 2c_1 + 2$. Solving this simple system of linear equations shows us that

$\displaystyle c_1 = 0; c_2 = 1$

Leaving us with a final answer of $y = e^{2t}\cos(2t)$ $\displaystyle y = e^{2t}\cos(2t)$

(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)

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