Y''+y'+y=1
solve ordinary differential equation

y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1

Step-by-step explanation:

y" + y' + y = 1

This is a second order nonhomogenous differential equation with constant coefficients.

First, find the roots of the complementary solution.

y" + y' + y = 0

r² + r + 1 = 0

r = [ -1 ± √(1² − 4(1)(1)) ] / 2(1)

r = [ -1 ± √(1 − 4) ] / 2

r = -1/2 ± i√3/2

These roots are complex, so the complementary solution is:

y = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t)

Next, assume the particular solution has the form of the right hand side of the differential equation.  In this case, a constant.

y = c

Plug this into the differential equation and use undetermined coefficients to solve:

y" + y' + y = 1

0 + 0 + c = 1

c = 1

So the total solution is:

y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1

To solve for c₁ and c₂, you need to be given initial conditions.

the correct answers i think are b. and d.

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