MATH SOLVE

6 months ago

Q:
# Y''+Y'+Y=1solve ordinary differential equation

Accepted Solution

A:

Answer:y(t) = cβ e^(-1/2 t) cos(β3/2 t) + cβ e^(-1/2 t) sin(β3/2 t) + 1Step-by-step explanation:y" + y' + y = 1This is a second order nonhomogenous differential equation with constant coefficients.First, find the roots of the complementary solution.y" + y' + y = 0rΒ² + r + 1 = 0r = [ -1 Β± β(1Β² β 4(1)(1)) ] / 2(1)r = [ -1 Β± β(1 β 4) ] / 2r = -1/2 Β± iβ3/2These roots are complex, so the complementary solution is:y = cβ e^(-1/2 t) cos(β3/2 t) + cβ e^(-1/2 t) sin(β3/2 t)Next, assume the particular solution has the form of the right hand side of the differential equation. Β In this case, a constant.y = cPlug this into the differential equation and use undetermined coefficients to solve:y" + y' + y = 10 + 0 + c = 1c = 1So the total solution is:y(t) = cβ e^(-1/2 t) cos(β3/2 t) + cβ e^(-1/2 t) sin(β3/2 t) + 1To solve for cβ and cβ, you need to be given initial conditions.