Question

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.

Answer 1

`y=4e^(-(x+1))` will be the solutions.

Explanation:

The given equation is
`y=C_(1)e^(x)+C_(2)e^(-x)`

Therefore, for x = -1

`4=C_(1)e^(-1)+C_(2)e^(1)` ------(1)

Now y'(-1) = -4

y'(x) =
`C_(1)e^(x)-C_(2)e^(-x)` = -4

`C_(1)e^(-1)-C_(2)e^(1)` = -4 -----(2)

By adding equation (1) and (2)

`2C_(1)e^(-1)=0`

`C_(1)=0`

From equation (1),

`4=0+C_(2)e^(1)`

`C_(2)=4e^(-1)`

By placing the values in the parent equation

y =
`4e^(-1)* e^(-x)`

y =
`4e^(-(x+1))`