Question

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

Answer 1

Answer:

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

Step-by-step explanation:

Given

$y=c_1e^x +c_2e^{-x$

$y(1) = 0$

$y'(1) =e$

Required

The solution

Differentiate $y=c_1e^x +c_2e^{-x$

$y' = c_1e^x - c_2e^{-x}$

Next, we solve for c1 and c2

$y(1) = 0$ implies that; x = 1 and y = 0

So, we have:

$y=c_1e^x +c_2e^{-x$

$0 = c_1 * e^1 + c_2 * e^{-1}$

$0 = c_1 e + \frac{1}{e}c_2$ --- (1)

$y'(1) =e$ implies that: x = 1 and y' = e

So, we have:

$y' = c_1e^x - c_2e^{-x}$

$e = c_1 * e^1 - c_2 * e^{-1}$

$e = c_1 e - \frac{1}{e}c_2$ --- (2)

Add (1) and (2)

$0 + e = c_1e + c_1e + \frac{1}{e}c_2 - \frac{1}{e}c_2$

$e = 2c_1e$

Divide both sided by e

$1 = 2c_1$

Divide both sides by 2

$c_1 = \frac{1}{2}$

Substitute $c_1 = \frac{1}{2}$ in $0 = c_1 e + \frac{1}{e}c_2$

$0 = \frac{1}{2} e+ \frac{1}{e}c_2$

Rewrite as:

$\frac{1}{e}c_2 = -\frac{1}{2} e$

Multiply both sides by e

$c_2 = -\frac{1}{2} e^2$

So, we have:

$y=c_1e^x +c_2e^{-x$

$y = \frac{1}{2}e^x -\frac{1}{2} e^2 * e^{-x}$

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$