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.