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Chapter 4: Problem 71

Find the solution of the following initial value problems. $$f^{\prime}(u)=4(\cos u-\sin 2 u) ; f(\pi / 6)=0$$

Short Answer

Expert verified
Question: Find the solution to the given initial value problem: $$f^{\prime}(u)=4(\cos u-\sin 2 u)$$ with $$f(\pi / 6)=0$$. Answer: The solution to the given initial value problem is $$f(u) = 4\left( \sin u + \frac{1}{2}\cos 2u \right) - 3$$.

Step by step solution

01

Rewrite the differential equation

Write the given differential equation as: $$\frac{df(u)}{du} = 4(\cos u - \sin 2u)$$
02

Integrate both sides

Integrate both sides of the equation with respect to $$u$$: $$\int \frac{df(u)}{du} du = \int 4(\cos u - \sin 2u) du$$ This simplifies to: $$f(u) = 4 \int (\cos u - \sin 2u) du$$
03

Integrate the function

Now, let's break down the integral of the function and integrate each term: $$f(u) = 4 \left( \int \cos u du - \int \sin 2u du \right)$$ Using the integration rules, we get: $$f(u) = 4\left( \sin u + \frac{1}{2}\cos 2u \right) + C$$ where $$C$$ is the constant of integration.
04

Apply the initial condition to find the constant

Now, use the initial condition $$f(\pi / 6) = 0$$ to find the value of $$C$$: $$0 = 4\left( \sin(\pi / 6) + \frac{1}{2}\cos (\pi / 3) \right) + C$$ Simplify and solve for $$C$$: $$0 = 4\left( \frac{1}{2} + \frac{1}{4} \right) + C$$ $$0 = 2 + 1 + C$$ Hence, $$C = -3$$.
05

Write down the solution

Now, substitute the value of $$C$$ into the formula for $$f(u)$$ to obtain the solution to the initial value problem: $$f(u) = 4\left( \sin u + \frac{1}{2}\cos 2u \right) - 3$$ This is the solution to the given initial value problem.

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