Question

In Exercises \(21-24\) , solve the initial value problem using the Fundamental Theorem. (Your answer will contain a definite integral.) \(\frac{d u}{d x}=\sqrt{2+\cos x}\) and \(u=-3\) when \(x=0\)

Short answer

The solution to the differential initial value problem is \(u(x) = \int{\sqrt{2 + \cos(x)}} dx - 3\). The integral represents the antiderivative of the function \(\sqrt{2 + \cos(x)}\) and \(C = -3\) is the final constant obtained by considering the initial condition.

Step-by-step solution

Understanding the equation

The differential equation \(\frac{d u}{d x}=\sqrt{2+\cos x}\) represents the rate of change of the function u with respect to x. The task is to find the function u that satisfies this relationship and also fulfils the initial condition \(u(-3) = 0\)

Integrate both sides of the differential equation

Applying the fundamental theorem of calculus to the given differential equation, which states that the integral of the derivative of a function is the function itself, we get: \[u(x) = \int{\sqrt{2 + \cos(x)} dx} + C\], where C is the constant of integration and needs to be found by using the given initial condition.

Apply initial condition

According to the problem, when \(x = 0\), \(u = -3\). Plug these values into the equation as it is unknown how to solve the integral of \(\sqrt{2+\cos x}\) analytically. So, the result becomes: \[-3 = \int_{0}^{0}\sqrt{2 + \cos(x)} dx + C\]Solving this equation, we get \(C = -3\)

Final Solution

Substitute \(C = -3\) into the equation to get the final solution: \[u(x) = \int{\sqrt{2 + \cos(x)}} dx - 3\]. This is the function that satisfies the given differential and initial value problem.