Fahd King University of Petroleum & Minerals Department of Mathematics & Statistics Solution Math 202 Major Exam II The First Semester of 2009-2010 (091) Time Allowed: 90 Minutes
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Mobiles and calculators are not allowed in this exam. Write all steps clear.
Question # 1 2 3 4 5 6 Total
Marks
Maximum Marks 12 8 12 10 12 12 66
Solution of MATH 202 EXAM 2 (Term 091) Q:1 Consider the dierential equation y 00 5y 0 + 6y = 3e4x :
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(1)
Let y1 = e2x be a solution of the associated homogeneous equation of (1). (a) (6 points) Use method of reduction of order to nd a FIRST order SEPARABLE equation for the associated homogeneous equation of (1). Sol: Let y2 (x) = u (x) y1 (x) be the second solution of the associated homogeneous equation of (1).
0 0 00 0 0 00 Then y2 = u0 y1 + uy1 and y2 = u00 y1 + u0 y1 + u0 y1 + uy1 00 0 Substituting y2 ; y2 ; and y2 in the associated homogeneous equation of (1), we get, 00 0 u00 y1 + 2u0 y1 + uy1 0 ) u00 y1 + 2u0 y1
5u0 y1
0 5uy1 + 6uy1 = 0 0 5y1 + 6y1 ) = 0 0 5y1 + 6y1 = 0 and y1 (x) = e2x :
00 5u0 y1 + u(y1
) u00 e2x + 2u0 2e2x ) (u00 + 4u0
00 5u0 e2x = 0; since y1
5u0 ) e2x = 0 ) u00
u0 = 0; since e2x 6= 0
Let u0 = w; then u00 = w0 : Substituting in u00 u0 = 0 we get the rst order separable equation w0 w = 0:
(b) (3 points) Use the rst order separable equation obtained in part (a) to nd a second solution of the associated homogeneous equation of (1). Sol: The rst order separable equation w0 is ln jwj = x or w (x) = ex : Now u0 (x) = w (x) ) u (x) = w = 0 can be written as dw = dx whose solution w
The second solution of the associated homogeneous equation is y2 (x) = ex e2x = e3x (c) (3 points) Find a particular solution of (1) and write its general solution. Sol: Let yp = Ae4x be the particular solution of (1). 3 20Ae4x + 6Ae4x = 3e4x ) A = : 2 3 The general solution of (1) is y = C1 e2x + C2 e3x + e4x : 2 Then substituting in (1), we get 16Ae4x
R
w (x) dx =
R
ex dx = ex
Solution of MATH 202 EXAM 2 (Term 091)
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Q:2 (a) (3 points) The auxiliary equation of an 8th -order linear homogeneous DE with real coe cients has the roots m1 = 3 of multiplicity 1; m2 = 2 of multiplicity 3, and m3 = 3 + 2i of multiplicity 2. Write the general solution of the DE. Sol: The roots of the auxiliary equation are m1 = 3; m2 = 2; 2; 2 and m3 = 3 + 2i; 3 + 2i: 2i; 3 2i:
Since the auxiliary equation has real coe cients, the other roots are m = 3 The solution of 8th order DE is y = C1 e
3x
+C2 e2x +C3 xe2x +C4 x2 e2x +e3x (C5 cos (2x) + C6 sin (2x))+xe3x (C7 cos (2x) + C8 sin (2x))
(b) (5 points) Solve the following dierential equation y 000 y = 0: 1 = 0 ) (m 1) (m2 + m + 1) = 0
Sol: The auxiliary equation of this equation is m3 p 3i 1 The roots are m = 1; 2 2 The solution is y = C1 ex + e
1 x 2
C2 cos
p
3 x 2
+ C3 sin
p
3 x 2
Q:3 (12 points) Use ANNIHILATOR approach to nd the general solution of y 00 2y 0 + 5y = ex cos x + 2ex sin x:
Sol: Auxiliary equation of the associated homogeneous equation is m2 2m + 5 = 0 ) m = 1 2i:
The complementary function is yc = ex (C1 cos (2x) + C2 sin (2x)) : Given equation can be written in operator for as (D2 Applying Annihilator (D2 (D2 2D + 2) (D2 2D + 5) y = ex cos x + 2ex sin x
2D + 2) on both sides we get 2D + 2) (ex cos x + 2ex sin x) = 0 2m + 2) (m2 2m + 5) = 0
2D + 5) y = (D2
Auxiliary equation of this equation is (m2 )m=1 i; 1 2i
) y = ex (C1 cos (2x) + C2 sin (2x)) + ex (C3 cos (x) + C4 sin (x))
Solution of MATH 202 EXAM 2 (Term 091) Let yp = ex (A cos (x) + B sin (x))
0 then yp = ex (A cos (x) + B sin (x)) + ex ( A sin (x) + B cos (x)) 00 and yp = ex (A cos (x) + B sin (x)) + ex ( A sin (x) + B cos (x)) +ex ( A sin (x) + B cos (x)) + ex ( A cos (x) B sin (x))
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= 2ex ( A sin (x) + B cos (x)) Substituting in the given equation we get 2ex ( A sin (x) + B cos (x)) 2ex (A cos (x) + B sin (x))
2ex ( A sin (x) + B cos (x)) + 5ex (A cos (x) + B sin (x)) = ex cos x + 2ex sin x 1 2 and B = : 3 3 1 2 The general solution is y = yc +yp = ex (C1 cos (2x) + C2 sin (2x))+ex cos x + ex sin x 3 3 ) 3ex A cos (x) + 3ex C4 sin (x) = ex cos x + 2ex sin x ) A = Q:4 (10 points) Find the general solution of x2 y 00 + xy 0 + x2 given that y1 = x
1 2
1 4
y = x2 ;
3
cos x is a solution of the corresponding homogeneous equation. 1 4x2 y=x
1 2
1 Sol: The given equation can be written as y 00 + y 0 + 1 x Z
The second solution the of associated homogeneous equation is
R1 R Z Z 1 1 e P (x)dx e x dx e ln x 2 cos x 2 cos x y2 (x) = y1 (x) =x =x dx x 1 cos2 x x 1 cos2 x (y1 (x))2 Z Z 1 1 1 x1 = x 2 cos x dx = x 2 cos x sec2 xdx = x 2 cos x tan x = x x 1 cos2 x
1 2
sin x:
Now we nd yp using variation of parameters method W= x 1 x 2 0 x
1 2 3 2 1 2
cos x x x
1 2 1 2
x sin x 1 x 2
3 2
1 2
sin x
1 2
cos x
sin x + x
cos x
=x
1
W1 =
sin x
1 2
1 x 2 x
1 2
3 2
sin x + x
cos x 0
=
x
1
sin x
W2 =
cos x x
1 2
1 x 2
3 2
cos x
sin x x
1 2
=x
1
cos x
Solution of MATH 202 EXAM 2 (Term 091) Z Z
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u1 (x) =
yp = u1 (x) y1 (x) + u2 (x) y2 (x) = x 2 cos2 x + x 2 sin2 x = x The general solution is y = C1 x
1 2
R W1 dx = W
sin xdx = cos x and u2 (x) =
1 1
R W2 dx = cos dx = sin x W
1 2 1 2
cos x + C2 x
1 2
sin x + x
:
Q:5 (a) (8 points) Solve the following dierential equation 2x2 y 00 + 7xy 0 + 3y = 2x3 : Sol: To nd yc ; let y = xm ; then y 0 = mxm
1
and y 00 = m (m
1) xm
2
Substituting in the associated homogeneous equation, we get the auxiliary equation 2m (m yc = C1 x W= u1 (x) = 1) + 7m + 3 = 0 ) 2m2 + 5m + 3 = 0 ) (2m + 3) (m + 1) = 0 ) m =
1
1;
3 : 2
+ C2 x
1 2
3 2
: Let y1 = x = 1 x 2
1
and y2 = x 0 ; W1 = x
3 2
: x2 3 x 2Z
3 5 2
x
R W1 1 W2 dx = 2 x3 dx = x4 and u2 (x) = dx = W 2 W 43 1 1 x = x3 yp = u1 (x) y1 (x) + u2 (x) y2 (x) = x3 2 9 18 1 3 The general solution is y = C1 x 1 + C2 x 2 + x3 : 18
x Z
x2 3 x 2
3
5 2
7 2
=
x
1 2;
W2 =
x1 0 x2 x 49 x2 9
=1
R7 2 x 2 dx =
(b) (4 points)Transform the equation in part (a) into an equation with constant coe cients. Sol: Let t = ln x; then using chain rule we can write dy dy dt 1 dy = = dx dt dx x dt 2 dy d dy d = = dx2 dx dx dx
1 dy x dt
=
1 dy 1 d2 y dt + = x2 dt x dt2 dx
1 dy 1 d2 y 1 + x2 dt x dt2 x
Substituting in the equation 2x2 y 00 + 7xy 0 + 3y = 2x3 ; we get 1 d2 y 1 dy 1 dy + 7x + 3y = 2e3t 2 dt2 2 dt x x x dt d2 y dy dy ) 2 2 2 + 7 + 3y = 2e3t dt dt dt 2 dy dy ) 2 2 + 5 + 3y = 2e3t dt dt 2x2
Solution of MATH 202 EXAM 2 (Term 091) Q:6 (12 points) Use POWER SERIES method to solve the initial value problem y 00
1 X n=0
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2xy 0 + 8y = 0;
1 X n=1
y (0) = 3; y 0 (0) = 0:
1 X n=2
Sol: Let y =
1 X n=2
cn x ; then y =
2
n
0
ncn x
n1
and y = cn xn = 0
00
n (n
1) cn xn
2
n (n
1 X n=3
1) cn xn n (n
1 X k=1 1 X k=1
2
1 X n=1
ncn xn + 8 2
1 X n=1
1 X n=0 n
2c2 +
1) cn x
n2
ncn x + 8c0 + 8
k
2c2 + 8c0 + 2c2 + 8c0 +
(k + 2) (k + 1) ck+2 x
2
1 X k=1
1 X n=1 k
cn xn = 0
1 X k=1
kck x + 8
ck xk = 0
(k + 2) (k + 1) ck+2 xk + (8 4c0 and ck+2 = c1
2k ) ck xk = 0
2c2 + 8c0 = 0 ) c2 = 6 c1 32 4 k = 2; c4 = c2 43 2 c3 k = 3; c5 = 54 0 k = 4; c6 = c4 65 k = 1; c3 = = =
(2k 8) ck ; k = 1; 2; 3; ::: (k + 2) (k + 1)
1 4c0 ( 4c0 ) = 3 3 1 = c1 10 = 0: So c2m = 0 for m = 3; 4; ::: 4 4x2 + x4 +c1 x 3 x3 + 15 x + ::: 10
y = c0 +c1 x+c2 x2 ++c3 x3 +c4 x4 +c5 x5 +c6 x6 ::: = c0 1 y 0 = c0 8x + 16 3 x 3 + c1 1 1 3x2 + x4 + ::: 2
y (0) = 3 ) c0 = 3 and y 0 (0) = 0 ) c1 = 0: The solution is y = 3 1 4 4x2 + x4 3 =3 12x2 + 4x4 :
Solution of Math 202 - 091 Exam 2 using Maple
> restart: with(DEtools):
Q:1
> Ode1:=diff(y(x),x$2)-5*diff(y(x),x$1)+6*y(x)=3*exp(4*x); 2 (4 x ) d d Ode1:= 2 y( x) 5 y( x) + 6 y( x) = 3e dx dx
> dsolve(Ode1);
y( x) = e
(2 x )
_C2 + e
(3 x )
3 (4 x ) _C1 + e 2
Q:2(b)
> Ode2:=diff(y(x),x$3)-y(x)=0; 3 d Ode2:= 3 y( x) y( x) = 0 dx > dsolve(Ode2);
y( x) = _C1ex + _C2e
x 2 2 3 x 3 x + _C3e cos sin 2 2 x
Q:3
> Ode3:=diff(y(x),x$2)2*diff(y(x),x$1)+5*y(x)=exp(x)*cos(x)+2*exp(x)*sin(x); 2 d d Ode3:= 2 y( x) 2 y( x) + 5 y( x) = ex cos( x) + 2 ex sin( x) dx dx > dsolve(Ode3);
1 y( x) = ex sin( 2 x) _C2 + ex cos( 2 x) _C1 + ex ( cos( x) + 2 sin( x) ) 3
Q:4
> Ode4:=x^2*diff(y(x),x$2)+x*diff(y(x),x$1)+(x^21/4)*y(x)=x^(3/2); 2 (32) / d 1 2 d Ode4:=x 2 y( x) + x y( x) + x2 y( x) = x dx dx 4 > dsolve(Ode4);
y( x) =
sin x) _C2 cos( x) _C1 1 ( + + x x x
Q:5(a)
> Ode5:=2*x^2*diff(y(x),x$2)+7*x*diff(y(x),x$1)+3*y(x)=2*x^3; 2 d d Ode5:=2 x2 2 y( x) + 7 x y( x) + 3 y( x) = 2 x3 dx dx > dsolve(Ode5);
_C2 _C1 x3 y( x) = (3/2) + + x 18 x
Q:6
> Ode6:=diff(y(x),x$2)-2*x*diff(y(x),x$1)+8*y(x)=0; 2 d d Ode6:= 2 y( x) 2 x y( x) + 8 y( x) = 0 dx dx > dsolve({Ode6,y(0)=3,D(y)(0)=0}); y( x) = 3 12x2 + 4 x4