Question

: To solve the initial value problem u′′(x)=12e2x−64e−4x with u(0)=1 and u′(0)=5, you can solve the differential equation and apply the initial conditions to determine the particular solution.

Answer 1

To solve the initial value problem
`u′′(x)=12e2x−64e−4x` with
`u(0)=1`and
`u′(0)=5` we can solve the differential equation and apply the initial conditions to determine the particular solution.

Step-by-step explanation:

To solve the initial value problem
`u′′(x) = 12e2x − 64e−4x with u(0) = 1` and
`u′(0) = 5` we can solve the differential equation and apply the initial conditions to determine the particular solution.

First, we solve the dwithfferential equation:

1. Let
   `y = u′(x).`
2. This means
   `y′ = u′′(x).`
3. So, we have
   `y′ = 12e2x − 64e−4x.`
4. To solve for
   `y` we integrate both sides of the equation with respect to
   `x`.
5. This gives us
   `y = ∫(12e2x − 64e−4x)dx.`
6. Integrating the right side of the equation,
7. we get
   `y = 6e2x + 16e−4x + C₁,`
8. where
   `C₁` is the constant of integration.

Next, we apply the initial conditions:

1. We know that
   `u(0) = 1`and
   `u′(0) = 5`.
2. Substituting
   `x = 0` into the equation for u
3. we get
   `u(0) = 6e^0 + 16e^0 + C₁ = 6 + 16 + C₁ = 22 + C₁ = 1.`
4. From this, we can determine that
   `C₁ = -21.`

Therefore, the particular solution to the initial value problem is,

`u(x) = 6e2x + 16e−4x - 21.`