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)

answer

a,b, and d

explanation

a reflection angle is more than

but less than;

an angle that measures

is a reflex angle.(true)

an angle that measures less than

is acute.(true)

provided we are considering angles greater or equal to zero.

an acute angle lies in the range.

an obtuse angle is less than 180° but more than 90°

an angle that measures 300° is obtuse (false)

an angle that measures 65° is an acute angle.(true)

the answer is d

the answer is in the first quadrant, which make the y-int positive.

since the dots are going down, therefore the correlation is negative

i hope it is right.