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Solve the equation y′′ 361y=e2x where y(0)=y′(0)=0. y(x)=

User Lesli
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Final answer:

The differential equation y'' + 361y = e^2x with initial conditions y(0)=0 and y'(0)=0 is solved by combining the complementary solution of the homogeneous equation y'' + 361y = 0 and a particular solution found using the method of undetermined coefficients.

Step-by-step explanation:

To solve the differential equation y'' + 361y = e2x with the initial conditions y(0)=0 and y'(0)=0, we start by finding the complementary solution yc of the homogeneous equation y'' + 361y = 0. This is a second-order differential equation with constant coefficients. The characteristic polynomial is r2 + 361 = 0, which has roots r = ±19i. Thus, the complementary solution is yc = A √ cos(19x) + B √ sin(19x).

To find the particular solution yp, we use the method of undetermined coefficients. The right-hand side of the given equation is e2x, so we guess a particular solution of the form yp = Cxe2x. Substituting yp into the original equation and solving for C gives us the particular solution.

Combining the complementary and particular solutions gives the general solution y(x) = yc + yp. Applying the initial conditions, we solve for A and B and find the specific solution that satisfies these conditions.

User Vickisys
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