Modeling Higher-Order Differential Equations - Differential Equations

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Question

For the initial value problem

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

Which if the following intervals containing do NOT guarantees the existence of a unique solution?

Answer

Putting the equation in standard form we get that

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

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

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

Compare your answer with the correct one above

Question

For the initial value problem

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

Which if the following intervals containing do NOT guarantees the existence of a unique solution?

Answer

Putting the equation in standard form we get that

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

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

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

Compare your answer with the correct one above

Question

For the initial value problem

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

Which if the following intervals containing do NOT guarantees the existence of a unique solution?

Answer

Putting the equation in standard form we get that

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

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

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

Compare your answer with the correct one above

Question

For the initial value problem

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

Which if the following intervals containing do NOT guarantees the existence of a unique solution?

Answer

Putting the equation in standard form we get that

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

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

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

Compare your answer with the correct one above

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