Author Topic: TUT5103 Quiz4  (Read 488 times)

Che Liang

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TUT5103 Quiz4
« 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 r1 = -2, r2 = -2.

Since the characteristic equation has repeated roots,
the general solution of the differential equation is,
y(t) = c1e-2t + c2te-2t        (1)   
cc1 and cc2 are constants.

And y’(t) = -2c1e-2t + c2(-2te-2t +e-2t)      (2)

Sub y(-1) = 2, y’(-1) = 1 into (1) and (2).

Get 2 = c1e2 - c2e2      (3)
And 1 = -2c_{1}e\{2} + c_{2}(3\e^{2})      (4)

Combine (3) and (4)
We can get c1 = 7e-2 and c1 = 5e-2

Sub c1 and c2 into (1),
We can get y(t) = 7e-2e-2t + 5e-2te-2t
Hence, the general solution of given initial value is,
 y(t) = 7e-2(t+1)+ 5te-2(t+1).