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lisov135 [29]
2 years ago
13

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with

initial values y(0) = −2 and y′(0) = 1.
Write the form of the particular solution and its derivatives. (Use A, B, C, etc. for undetermined coefficients.
Y =
Y' =
Y" =
Mathematics
1 answer:
Zarrin [17]2 years ago
7 0

First check the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -2, so the characteristic solution is

y_c = C_1e^{-2t} + C_2te^{-2t}

For the particular solution corresponding to 12te^{-2t}, we might first try the <em>ansatz</em>

y_p = (At+B)e^{-2t}

but e^{-2t} and te^{-2t} are already accounted for in the characteristic solution. So we instead use

y_p = (At^3+Bt^2)e^{-2t}

which has derivatives

{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}

{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}

Substituting these into the left side of the ODE gives

(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.

For the second solution corresponding to -8t-12, we use

y_p = Ct + D

with derivative

{y_p}' = C

{y_p}'' = 0

Substituting these gives

4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12

so that 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -1.

Then the general solution to the ODE is

y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1

Given the initial conditions <em>y</em> (0) = -2 and <em>y'</em> (0) = 1, we have

-2 = C_1 - 1 \implies C_1 = -1

1 = -2C_1 + C_2 - 2 \implies C_2 = 1

and so the particular solution satisfying these conditions is

y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1

or

\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}

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