Respuesta :
Answer:
c₁ = 1/2
c₂ = - e²/2
y = (1/2)*(eˣ - e²⁻ˣ)
Step-by-step explanation:
Given
y = c₁eˣ + c₂e⁻ˣ
y(1) = 0
y'(1) = e
We get y' :
y' = (c₁eˣ + c₂e⁻ˣ)' ⇒ y' = c₁eˣ - c₂e⁻ˣ
then we find y(1) :
y(1) = c₁e¹ + c₂e⁻¹ = 0
⇒ c₁ = - c₂/e² (I)
then we obtain y'(1):
y'(1) = c₁e¹ - c₂e⁻¹ = e (II)
⇒ (- c₂/e²)*e - c₂e⁻¹ = e
⇒ - c₂e⁻¹ - c₂e⁻¹ = - 2c₂e⁻¹ = e
⇒ c₂ = - e²/2
and
c₁ = - c₂/e² = - (- e²/2) / e²
⇒ c₁ = 1/2
Finally, the equation will be
y = (1/2)*eˣ - (e²/2)*e⁻ˣ = (1/2)*(eˣ - e²⁻ˣ)
Applying the initial conditions, it is found that the solution is:
[tex]y = \frac{1}{2}e^{x} - \frac{e^2}{2}e^{-x}[/tex]
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The solution for the PVI is given by:
[tex]y = c_1e^{x} + c_2e^{-x}[/tex]
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The condition [tex]y(1) = 0[/tex] means that when [tex]x = 0, y = 1[/tex], and thus, we get:
[tex]c_1e + c_2e^{-1} = 0[/tex]
[tex]c_1e+ \frac{c_2}{e} = 0[/tex]
[tex]c_1e^{2} + c_2 = 0[/tex]
[tex]c_2 = -c_1e^{2}[/tex]
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The derivative is:
[tex]y^{\prime}(x) = c_1e^{x} - c_2e^{-x}[/tex]
Applying the condition [tex]y^{\prime}(1) = e[/tex], we get:
[tex]c_1e - \frac{c_2}{e} = e[/tex]
Considering [tex]c_2 = -c_1e^{2}[/tex]:
[tex]c_1e + c_1\frac{e^2}{e} = e[/tex]
[tex]c_1e + c_1e = e[/tex]
[tex]2c_1e = e[/tex]
[tex]2c_1 = 1[/tex]
[tex]c_1 = \frac{1}{2}[/tex]
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The second constant is:
[tex]c_2 = -c_1e^{2} = -\frac{e^2}{2}[/tex]
And the solution is:
[tex]y = \frac{1}{2}e^{x} - \frac{e^2}{2}e^{-x}[/tex]
A similar problem is given at https://brainly.com/question/13244107
