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family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. 12. y(1) 0, y(1) e

Sagot :

MrRoyal

Answer:

[tex]y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}[/tex]

Step-by-step explanation:

Given

[tex]y=c_1e^x +c_2e^{-x[/tex]

[tex]y(1) = 0[/tex]

[tex]y'(1) =e[/tex]

Required

The solution

Differentiate [tex]y=c_1e^x +c_2e^{-x[/tex]

[tex]y' = c_1e^x - c_2e^{-x}[/tex]

Next, we solve for c1 and c2

[tex]y(1) = 0[/tex] implies that; x = 1 and y = 0

So, we have:

[tex]y=c_1e^x +c_2e^{-x[/tex]

[tex]0 = c_1 * e^1 + c_2 * e^{-1}[/tex]

[tex]0 = c_1 e + \frac{1}{e}c_2[/tex] --- (1)

[tex]y'(1) =e[/tex] implies that: x = 1 and y' = e

So, we have:

[tex]y' = c_1e^x - c_2e^{-x}[/tex]

[tex]e = c_1 * e^1 - c_2 * e^{-1}[/tex]

[tex]e = c_1 e - \frac{1}{e}c_2[/tex] --- (2)

Add (1) and (2)

[tex]0 + e = c_1e + c_1e + \frac{1}{e}c_2 - \frac{1}{e}c_2[/tex]

[tex]e = 2c_1e[/tex]

Divide both sided by e

[tex]1 = 2c_1[/tex]

Divide both sides by 2

[tex]c_1 = \frac{1}{2}[/tex]

Substitute [tex]c_1 = \frac{1}{2}[/tex] in [tex]0 = c_1 e + \frac{1}{e}c_2[/tex]

[tex]0 = \frac{1}{2} e+ \frac{1}{e}c_2[/tex]

Rewrite as:

[tex]\frac{1}{e}c_2 = -\frac{1}{2} e[/tex]

Multiply both sides by e

[tex]c_2 = -\frac{1}{2} e^2[/tex]

So, we have:

[tex]y=c_1e^x +c_2e^{-x[/tex]

[tex]y = \frac{1}{2}e^x -\frac{1}{2} e^2 * e^{-x}[/tex]

[tex]y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}[/tex]

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