In this problem, y = c1ex + c2e−x is a two-parameter family of solutions of the second-order de y'' − y = 0. find c1 and c2 given the following initial conditions. (your answers will not contain a variable.) y(1) = 0, y'(1) = e c1 = incorrect: your answer is incorrect. c2 = incorrect: your answer is incorrect. find a solution of the second-order ivp consisting of this differential equation and the given initial conditions. y = incorrect: your answer is incorrect.

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)

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

∠cbe and ∠deb are same-side interior angles.

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