In this problem, y = c₁eˣ + c₂e⁻ˣ is a two-parameter family of solutions of the second-order DE y'' − y = 0. Find a solution of the second-order IV P consisting of this differential equation and the given initial conditions. y(-1) = 4, y'(-1) = -4.

Respuesta :

Answer:

[tex]y=4e^{-(x+1)}[/tex] will be the solutions.

Step-by-step explanation:

The given equation is [tex]y=C_{1}e^{x}+C_{2}e^{-x}[/tex]

Therefore, for x = -1

[tex]4=C_{1}e^{-1}+C_{2}e^{1}[/tex] ------(1)

Now y'(-1) = -4

y'(x) = [tex]C_{1}e^{x}-C_{2}e^{-x}[/tex] = -4

[tex]C_{1}e^{-1}-C_{2}e^{1}[/tex] = -4 -----(2)

By adding equation (1) and (2)

[tex]2C_{1}e^{-1}=0[/tex]

[tex]C_{1}=0[/tex]

From equation (1),

[tex]4=0+C_{2}e^{1}[/tex]

[tex]C_{2}=4e^{-1}[/tex]

By placing the values in the parent equation

y = [tex]4e^{-1}\times e^{-x}[/tex]

y = [tex]4e^{-(x+1)}[/tex]