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### Messages - Che Liang

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##### Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 09:29:16 AM »
y'' -6y' + 25y = 16e3x + 102sinx
y(0) = 0, y'(0) = 0

(a) The characteristic equation is r2 - 6r +25 = 0,
then (r-3)2 = -16, so r = 3 + 4i or 3- 4i.

so yc(x) = c1e3xcos4x + c2e3xsin4x

Think about y'' -6y' + 25y = 16e3x.
Let y1 = Ae3x,
so y1' = 3Ae3x, y1'' = 9Ae3x.
Substitute them in the equation,
we get 9Ae3x - 18Ae3x +25Ae3x = 16e3x.
So 9A-18A+25A = 16, A = 1.
We can get: y1 = e3x.

Think about y'' -6y' + 25y = 102sinx.
Let y2 = Bcosx + Csinx,
so y2' = -Bsinx +Ccosx,
y2'' = -Bcosx - Csinx.
Substitute them in the equation,
we get -Bcosx - Csinx +6Bsinx - 6Ccosx + 25Bcosx + 25Csinx = 102sinx.
So -B - 6C + 25B =0 and -C + 6B + 25C = 102. Then we know B = 1 and C =4.
We can get: y2 = cosx + 4sinx.

Therefore the general solution is:
y(x) = c1e3xcos4x + c2e3xsin4x + e3x + cosx + 4sinx

(b)
y' = c1(-4e3xsin4x + 3e3xcos4x)
+ c2(4e3xcos4x + 3e3xsin4x)
+ 3e3x - sinx + 4cosx
Substitute  y(0) = 0, y'(0) = 0,
We get 0 = c1 + 0 + 1 + 1 + 0, so c1 = -2.
And   0 = c1(-0 + 3) + c2(4 + 0) + 3 - 0 + 4
so 0 = -6 + 4c2 + 7，c2 = -1/4.

Therefore the solution is:
y(x) = -2e3xcos4x + (-1/4)e3xsin4x + e3x + cosx + 4sinx

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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)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).

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