Final answer:
The provided initial-value problem is a differential equation that can be solved using an integrating factor to express y(x) as an integral-defined function. The integrating factor is e^{2 \sin x}, and the final solution involves an integral with initial conditions applied to find the constant of integration.
Step-by-step explanation:
The student is tasked with expressing the solution to an initial-value differential equation problem using an integral-defined function. Specifically, the problem provided is a first-order linear differential equation:
\( 2 \frac{dy}{dx} + (4 \cos x)y = x, \quad y(0)=1 \)
To solve this, we first need to find the integrating factor: \( \mu(x) \). The integrating factor is given by:
\( \mu(x) = e^{\int P(x) dx} \)
where P(x) is the coefficient of y in the differential equation. In this case, P(x) = 2 \cos x. Integrating P(x) gives:
\( \mu(x) = e^{\int 2 \cos x \, dx} = e^{2 \sin x} \)
Using the integrating factor, the solution to the differential equation is given by:
\( y(x) \mu(x) = \int \mu(x) Q(x) dx + C \)
where Q(x) is the right-hand side of the original differential equation, which is x in this case. Finally, apply the initial condition y(0) = 1 to find the constant C, and express y(x) in terms of an integral.
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