Question

Find f. f ″(x) = 5e^x − 8 sin(x), f(0) = 3, f (/2) = 0

Answer 1

The function F is:
`F(x) = 5e^x + 8sin(x) - 5xe^((\pi/2)) - 2`

Given Information:

We start with a second-order linear differential equation:

`\[ f^(\prime\prime)(x) = 5e^x - 8\sin(x) \]`

Along with the following initial conditions:

`\[ f(0) = 3, f\left((\pi)/(2)\right) = 0 \]`

Integration to Find
`\(f^\prime(x)\)` :

To find the first derivative
`\(f^\prime(x)\)`, we integrate the given second derivative:

`\[ \int (5e^x - 8\sin(x)) \,dx = 5e^x + 8\cos(x) + C_1 \]`

Here,
`\(C_1\)` is the constant of integration.

Application of Initial Condition for
`\(f^\prime(x)\)` :

Using the initial condition
`\(f^\prime\left((\pi)/(2)\right) = 0\)`, we substitute
`\(x = (\pi)/(2)\)` into the expression for
`\(f^\prime(x)\)`:

`\[ 5e^{(\pi)/(2)} + 8\cos\left((\pi)/(2)\right) + C_1 = 0 \]`

Solving for
`\(C_1\)`, we get
`\(C_1 = -5e^{(\pi)/(2)}\)`.

Integration to Find
`\(f(x)\)` :

Integrating
`\(f^\prime(x)\)`, we obtain
`\(f(x)\)`:

`\[ \int (5e^x + 8\cos(x) - 5e^{(\pi)/(2)}) \,dx = 5e^x + 8\sin(x) - 5xe^{(\pi)/(2)} + C_2 \]`

Here,
`\(C_2\)` is the constant of integration.

Application of Initial Condition for
`\(f(x)\)` :

Using the initial condition
`\(f(0) = 3\)`, we substitute
`\(x = 0\)` into the expression for
`\(f(x)\)` :

`\[ 5 + 8\sin(0) - 5(0)e^{(\pi)/(2)} + C_2 = 3 \]`

Solving for
`\(C_2\)`, we get
`\(C_2 = -2\)`

Final Solution:

Combining the expressions for
`\(f(x)\)` and the determined constants, the solution to the differential equation with the given initial conditions is:

`\[ f(x) = 5e^x + 8\sin(x) - 5xe^{(\pi)/(2)} - 2 \]`