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]
