Question

y ′′ +2y ′ +5y=2sinx, with y(0)=1,y ′ (0)=0. solving the initial value problem. Find the phase angle and identify the steady state and transient solutions.

Answer 1

To solve the initial value problem, use the method of undetermined coefficients to find the complementary and particular solutions.

Step-by-step explanation:

To solve the initial value problem y'' + 2y' + 5y = 2sin(x) with y(0) = 1 and y'(0) = 0, we can use the method of undetermined coefficients. Start by finding the complementary solution by assuming y = e^rt, which leads to the characteristic equation r^2 + 2r + 5 = 0.

The roots of this equation are complex: r = -1 ± 2i. Therefore, the complementary solution is y_c(x) = e^-x[c1cos(2x) + c2sin(2x)].

For the particular solution, we assume y_p(x) = A sin(x) + B cos(x), substitute it into the equation, and solve for A and B. Once we find the particular solution, the general solution is y(x) = y_c(x) + y_p(x).