Question

Solve the differential equation. $$ f^{\prime \prime}(x)=e^{x}, f^{\prime}(0)=2, f(0)=5 $$

Short answer

The solution to the differential equation is \(f(x) = e^x + x + 4\)

Step-by-step solution

Find the first integral of the function

The first integral of \(f^{\prime\prime}(x)=e^x\) is \(\int f^{\prime\prime}(x) dx = \int e^x dx = e^x + C\), where C is a constant that will be determined.

Find the second integral of the function

The second integral of \(f'(x)=e^x + C\) is \(\int f'(x) dx = \int (e^x + C) dx = e^x+Cx+D\), where D is another constant that will be determined.

Apply the initial conditions to determine the constants

We have 2 initial conditions given in the problem: \(f'(0)=2\) and \(f(0)=5\). Substituting these into the functions for \(f'(x)\) and \(f(x)\), we get: \(f'(0)=2 => e^0 + C = 2 => C = 2 - 1 = 1\), and \(f(0)=5 => e^0+0+D = 5 => D = 5-1 = 4\). Thus C=1 and D=4.

Write down the final solution with the determined constants

Substituting C and D into \(f(x)\), we get the final solution: \(f(x) = e^x + x + 4\)