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Makena Cantrell
Mathematics
14 September, 12:14
Y''+Y'+Y=1
solve ordinary differential equation
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Dwayne Bradshaw
14 September, 12:42
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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.
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