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**Quiz-4 / 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)

r = -2, -2.

The roots are r

Since the characteristic equation has repeated roots,

the general solution of the differential equation is,

y(t) = c

cc

And y’(t) = -2c

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

Get 2 = c

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

Combine (3) and (4)

We can get c

Sub c

We can get y(t) = 7e

Hence, the general solution of given initial value is,

y(t) = 7e

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)}.