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Solve the initial value problem: y'' - y = 2e^x + 3e^(2x), y(0) = 0, y'(0) = 0.

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Final answer:

To solve the initial value problem, find the general solution to the homogeneous equation using the characteristic equation, then find a particular solution to the nonhomogeneous equation, and use the initial conditions to determine the constants.

Step-by-step explanation:

To solve the initial value problem y'' - y = 2e^x + 3e^(2x), with initial conditions y(0) = 0 and y'(0) = 0, we must find the general solution to the homogeneous equation y'' - y = 0 and a particular solution to the nonhomogeneous equation.

For the homogeneous part, the characteristic equation is r^2 - 1 = 0, which has solutions r = 1 and r = -1. Therefore, the general solution to the homogeneous equation is yh = C1e^x + C2e^(-x), where C1 and C2 are constants.

To find a particular solution yp to the nonhomogeneous equation, we can use the method of undetermined coefficients. We assume a form for yp which resembles the nonhomogeneous part, so yp = Ae^x + Be^(2x) where A and B will be determined by substituting yp into the original nonhomogeneous equation and equating coefficients.

Once yh and yp are found, we combine them for the general solution y = yh + yp. Then, we use the initial conditions to find the values of the constants C1, C2, A, and B.

User Tim Webster
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