1



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

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download