Find f.f '''(x) = cos(x),    f(0) = 2,    f '(0) = 6,    f ''(0) = 5f(x) =

The given differential equation is f.f '''(x) = cos(x). This is a third order non-linear differential equation.

Let's solve it step by step:

Step 1: First, we need to find the general solution of the homogeneous equation f.f '''(x) = 0. This equation is separable, so we can write it as f '''(x) / f(x) = 0. Integrating both sides, we get f ''(x) = C1, where C1 is the constant of integration.

Step 2: The general solution of the homogeneous equation is f(x) = C1*x + C2, where C2 is another constant of integration.

Step 3: Now, we need to find a particular solution of the non-homogeneous equation f.f '''(x) = cos(x). We can guess a solution of the form f(x) = Acos(x) + Bsin(x), where A and B are constants to be determined.

Step 4: Substituting this guess into the differential equation, we get (Acos(x) + Bsin(x)) * (-Asin(x) + Bcos(x)) = cos(x). Equating the coefficients of sin(x) and cos(x) on both sides, we get the system of equations -A^2 + B^2 = 1 and 2AB = 0. Solving this system, we find that A = 1, B = 0 is a solution.

Step 5: Therefore, the general solution of the differential equation is f(x) = C1*x + C2 + cos(x).

Step 6: Finally, we need to determine the constants C1 and C2 using the initial conditions f(0) = 2 and f '(0) = 6. Substituting these conditions into the general solution and its derivative, we get the system of equations C2 + 1 = 2 and C1 = 6. Solving this system, we find that C1 = 6, C2 = 1.

So, the solution of the differential equation satisfying the initial conditions is f(x) = 6x + 1 + cos(x).