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Marking Scheme of F.6 Applied Mathematics (II) Quiz (2009-10-29)
1. (1 ( y2)(xdy + ydx) = tan(xy)dy
[pic] 2M
[pic] 1M
[pic] + C1 1M
2ln|sin(xy)| = ln[pic] + 2C1 ( lnsin2(xy) = lnC[pic] 1M
sin2(xy) = C[pic] 1A
2. [pic] + 4y = 2cosxcos3x
Auxiliary equation : (( 2 + 4 = 0 , ( = (2i 1M
yc = Acos2x + Bsin2x 1A
RHS = 2cosxcos3x = cos4x + cos2x 1M
Hence let yp = Ccos4x + Dsin4x + x(Ecos2x + Fsin2x) 1M
[pic] = (4Csin4x + 4Dcos4x + Ecos2x + Fsin2x + x((2Esin2x + 2Fcos2x)
[pic] = (16Ccos4x ( 16Dsin4x ( 2Esin2x + 2Fcos2x ( 2Esin2x + 2Fcos2x + x((4Ecos4x ( 4Fsin2x)
[pic] + 4y
= (16Ccos4x ( 16Dsin4x ( 4Esin2x + 4Fcos2x + x((4Ecos4x ( 4Fsin2x) + 4Ccos4x + 4Dsin4x + 4x(Ecos2x + Fsin2x) 1M
= (12Ccos4x ( 12Dsin4x ( 4Esin2x + 4Fcos2x ( cos4x + cos2x
∴ [pic] 1M
C = ([pic] , F = [pic]
Thus, y = Acos2x + Bsin2x ([pic] + [pic] 2A
3. ydx + [x ( (1(xy)cosy]dy = 0
ydx = [(1(xy)cosy ( x]dy 1M
[pic] = [pic] ( xcosy ( [pic]
[pic] + (cosy + [pic])x = [pic] 1M
Integrating factor = exp[pic] = exp(siny + lny) = y[pic] 1M+1A
xy[pic] = [pic] 1M
= [pic] = [pic] + C 1A
4. eu = x + 1
eu[pic]= 1 ( [pic] = e(u 1A
[pic] = [pic] = e(u[pic] 1A
[pic]= e(u[pic] + [pic][pic]
= e(u[pic] ( [pic]e(u[pic] 1M
= e(u[pic]e(u ( [pic]e(ue(u
= e(2u ([pic] ( [pic]) 1A
Substitute the above into the original differential equation,
e2ue(2u ([pic] ( [pic]) ( 2eu (e(u[pic]) ( 10y = e4u
[pic] ( 3[pic] ( 10y = e4u --- (*) 1A
Auxiliary equation (2 ( 3( ( 10 = 0, hence ( = 5 or (2 1M
( yc = C1e5u + C2e(2u 1A
Take yp = Ae4u ( [pic] = 4Ae4u ( [pic]= 16Ae4u 1M
Substitute the above into (*),
(16 ( 3(4 ( 10)Ae4u ( e4u 1M
A = ([pic]; hence y = C1e5u + C2e(2u ( [pic]e4u = C1(x + 1)5 + C2(x + 1)(2 ( [pic](x + 1)4 1A
5. Let x = u + h, y = v + k where h and k are constants to be determined.
dx = du and dy = dv. 1M
The original differential equation will be
[pic]
Set [pic], on solving, h = 1, k = 1 and the differential equation will be 1M
[pic] = [pic] = [pic] --- (*) 1M
Let z =[pic], v = zu ( [pic] = z + u[pic]
(*) will be
z + u[pic] = [pic] 1M
u[pic] = [pic] = [pic] 1M
[pic] = ([pic] --- (**)
Consider
[pic] ( [pic] ( [pic] + [pic] 1M
4z + 2 ( A(z + 2) + B(4z ( 1) ( (A + 4B)z + 2A ( B
([pic], on solving A =[pic], B =[pic]. 1A
(**) will be
[pic] + [pic] = (ln|u|
[pic]ln|4z ( 1| + [pic]ln|z + 2| + ln|u| = C1 1A
ln|4z ( 1| + ln(z + 2)2 + ln|u3| = 3C1
(4z ( 1)(z + 2)2u3 = [pic] 1M
[pic]2u3 = C
[4(y ( 1) ( (x ( 1)][(y ( 1) + 2(x ( 1)]2 = C ( (4y ( x ( 3)(2x + y ( 3)2 = C 1A
END
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