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Solve the following IVP: cos(x)y' +sin(x)y=2cos^3
(x)sin(x)−1. Given y(/4)=3√2 .0≤x≤π/2


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Final answer:

To solve the given initial value problem (IVP), we can use an integrating factor to rewrite the equation as a more manageable form. Integrating both sides and solving for y, we can find the solution to the IVP.

Step-by-step explanation:

To solve the given initial value problem (IVP), we will use an integrating factor to solve the linear first-order differential equation. The integrating factor for the given equation is e^(∫sin(x)dx) = e^(-cos(x)). Multiplying both sides of the equation by this integrating factor, we can rewrite the equation as d/dx[e^(-cos(x))y] = 2cos^(3)(x)sin(x) - e^(-cos(x)).

Integrating both sides of the equation with respect to x, we get e^(-cos(x))y = ∫(2cos^(3)(x)sin(x) - e^(-cos(x)))dx.

Solving the integral on the right-hand side, we obtain e^(-cos(x))y = -cos^(4)(x) - x + C, where C is the constant of integration. Finally, dividing both sides by e^(-cos(x)), we find y = (-cos^(4)(x) - x + C)e^(cos(x)).

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