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.