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MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: etheliachoi on October 18, 2019, 10:47:11 PM

Title: Quiz 4 (TUT0601)
Post by: etheliachoi on October 18, 2019, 10:47:11 PM

Find the general solution of this equation:

y''+y'-6y = 12e^3t+12e^-2t

Find the homogenous solution:

y''+y'-6y=0

r²+r-6r=0

(r+3)(r-2)=0

r₁ = -3, r₂ = 2

y₁(t)= e^-3t, y₂(t)= e^2t

Then,

y(t) =c₁e^-3t + c₂e^2t

Now, consider

y''+y'-6y = 12e^3t

Want to find:

Y(t) = Ae^3t

Y'(t) = 3Ae^3t, Y''(t) = 9Ae^3t

Then,

9Ae^3t + 3Ae^3t - 6Ae^3t = 12e^3t

Thus, solving the equation, we get that:

A = 2 and Y(t) = 2e^3t

Now, consider

y''+y'-6y = 12e^-2t

Want to find:

Y(t) = Be^-2t

Repeat for the process for Y(t) and solve for B. Hence, we get that

B = -3 and Y(t) = -3e^-2t

Lastly,

y = c₁e^-3t + c₂e^2t + 2e^3t - 3e^-2t