Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz4 => Topic started by: Che Liang on October 18, 2019, 02:02:42 PM

y’’ + 4y’ + 4y = 0, y(1) = 2, y’(1) = 1.
The characteristic equation is r^2 + 4r + 4 = 0,
then (r + 2)^{2} = 0,
r = 2, 2.
The roots are r_{1} = 2, r_{2} = 2.
Since the characteristic equation has repeated roots,
the general solution of the differential equation is,
y(t) = c_{1}e^{2t} + c_{2}te^{2t} (1)
cc_{1} and cc_{2} are constants.
And y’(t) = 2c_{1}e^{2t} + c_{2}(2te^{2t} +e^{2t}) (2)
Sub y(1) = 2, y’(1) = 1 into (1) and (2).
Get 2 = c_{1}e^{2}  c_{2}e^{2} (3)
And 1 = 2c_{1}e\{2} + c_{2}(3\e^{2}) (4)
Combine (3) and (4)
We can get c_{1} = 7e^{2} and c_{1} = 5e^{2}
Sub c_{1} and c_{2} into (1),
We can get y(t) = 7e^{2}e^{2t} + 5e^{2}te^{2t}
Hence, the general solution of given initial value is,
y(t) = 7e^{2(t+1)}+ 5te^{2(t+1)}.