MAT244--2019F > Term Test 1

Problem 4 (morning)

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Che Liang:
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

Kunpeng Liu:
$$Let \, \, \, y=e^{rt}\, \,\, \, {y}'=re^{rt}\, \, \, {y}''=r^{2}e^{rt}\\\\Then\, \, \, \, \, \, r^{2}-6r+25=0\, \, \, \, \, \, \, \\\\\\\\\\r1=\frac{6+\sqrt{36-4\cdot 25\cdot 1}}{2}=3+4i\, \, \, \, \, \, \, r2=\frac{6-\sqrt{36-4\cdot 25\cdot 1}}{2}=3-4i\\\\Yc(x)=C1e^{3x}cos4x+C2e^{3x}sin4x\\\\\\\\\\Let\, \, Yp(x)=Ae^{3x}\, \, \,\, \, \, {y}'=3Ae^{3x}\, \, \,\, \, {y}''=9Ae^{3x}\\\\\\\\\\9Ae^{3x}-18Ae^{3x}+25Ae^{3x}=16e^{3x}\, \, \, \, \, \, 16Ae^{3x}=16e^{3x}\, \, \, \, \, 16A=16\, \,\, \, \, \, A=1\\\\\\\\\\Yp(x)=e^{3x}\\\\\\\\\\let \, \, \, \, Yq(x)=Asinx+Bcosx\, \, ,\, \, {y}'=Acosx-Bsinx\, \, ,\, \, {y}''=-Asinx-Bcosx\\\\\\\\\\-Asinx-Bcosx-6Acosx+6Bsinx+25Asinx+25Bcosx=102sinx\\\\\\\\\\(24A+6B)sinx+(24B-6A)cosx=102sinx\, \, \, \, \, \, 24A+6B=102\, \, \, \, 24B-6A=0\\\\A=4\, \, \, \, B=1\\\\\\\\\\Yq(x)=4sinx+cosx\\\\\\\\\\Y(x)=C1e^{3x}cos4x+C2e^{3x}sin4x+e^{3x}+4sinx+cosx\\\\y(0)=0\\\\\\because y(0)=C1+2=0\, \, \, \, C1=-2\\\\\\\\\\\because\, \, \, \, \, {y}'(0)=0\, \, \, \, {y}'(0)=3C1-0+0+4C2+3+4=0\, \, \, \, C2=-\frac{1}{4}\\\\\\\\\\Y(x)=-2e^{3x}cos4x-\frac{1}{4}e^{3x}sin4x+e^{3x}+4sinx+cosx$$

GuangyuDu:
Question 4:
$y''-6y'+25y=16e^{3x}+102\sin (x), y(0)=0, y'(0)=0$

Solution:
$\Gamma^2-6\Gamma+25=0, \Gamma=3\pm 4\pi$
$y_c(x)=c_1e^{3x}\cos (4x)+c_2e^{3x}\sin (4x)$
$y_p(x)=Ae^{3x},y_p'(x)=3Ae^{3x},y_p''(x)=9Ae^{3x}$
$y''-6y'+25y=16Ae^{3x}=16e^{3x}$
$A=1,y_p(x)=e^{3x}$
$y_p(x)=B\cos (x)+C\sin x, y_p'(x)=-B\sin x+C\cos x,$
$y_p''(x)=-B\cos x-C\sin x$
$y''-6y'+25y=102\sin x$
$B=1,C=4,y_p(x)=\cos x+4\sin x$
$y=y_c+y_p=c_1e^{3x}\cos (4x) +c_2e^{3x}\sin (4x)+e^{3x}+\cos x+4\sin x$

Plug in $y(0)=0,y'(0)=0$

we have $c_1=-2,c_2=-\frac14$
$y=-2e^{3x}\cos (4x)-\frac14e^{3x}\sin (4x)+e^{3x}+\cos x+4\sin x$