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ENGI 3424 Mid Term Test - Solutions 2020 October 15 1.

The displacement x ( t ) of the free end of a mass-spring system beyond the equilibrium position is governed by the ordinary differential equation d2 x dx + 4 + 40x = 36 e −2t 2 dt dt together with the initial conditions x( 0) = x (a) By any valid method, find the complete solution to this initial value problem. (b) Is the system under-damped, critically damped or overdamped?

(a) A.E.:

 2 + 4  + 40 = 0   =

Immediately: (b) the system is under -damped (a) C.F.:

xC = e

−2 t

−4  16 − 160 −4  − 144 − 4  12 j = = = − 2 6 j 2 2 2

(  is a complex conjugate pair)

( A cos 6t + B sin 6t )

P.S. by the method of undetermined coefficients: − − r( t) = 36 e−2 t is not a constant multiple of either x1 = e 2 t cos 6t or x2 = e 2t sin 6t Therefore try xP = c e −2t Substitute into 



 x P

( 4 − 8 + 40) c e−2t = 36 e− 2t

= 36 e−2t :  36c = 36  c = 1

−2t

xP = e

OR P.S. by the method of variation of parameters: x1 = e −2t cos 6t , x2 = e − 2t sin 6t , r( t) = 36 e−2 t

W =

x1 x

(

x2

=

e −2t cos 6t 2t

[12] [2]

( −2 cos 6t − 6sin 2t )

e −2t sin 6t e −2t ( −2sin 6t + 6 cos 2t )

)

= e −4t −2sin 6t cos 6t + 6 cos 2 2t + 2cos 6t sin 6t + 6sin 2 6t = 6 e −4t

ENGI 3424

Mid Term Test - Solutions

Page 2 of 10

1 (a) (continued) W1 = u

0

x2

r

x

v

−36 e− 4t sin 6t

W W

W2 = W W

(

x

)

= − 6sin 6 t  u = cos 6t

6 e − 4t

x1 0

)(

= − x r = − e −2t sin 6t 36 e −2t = − 36 e −4t sin 6t

(

= + x1r = e− 2t cos 6t

36 e−4t cos 6t 6 e −4 t

)( 36e− 2t ) = 36e− 4t cos 6t

= 6 cos 6t  v = sin 6t

(

)

(

)

xP = u  x1 + v  x2 = ( cos 6t ) e − 2t cos 6t + ( sin 6t ) e − 2t sin 6t = e − 2t [Clearly the method of undetermined coefficients is much faster!]

G.S.: 2t 2t 2t x (t ) = xC + xP = e − ( A cos 6t + B sin 6t ) + e − = e − (1 + A cos 6t + B sin 6t ) 

2t

x

( −2 − 2 Acos 6t − 2 B sin 6 t

I.C.: x (0 ) = 6  e 0 (1+ A + 0 )= 6 

x C.S.:



+ 0 − 6 Asin 6t + 6 B cos 6t)

A= 5

e 0 (− 2− 10− 0 + 0− 0+ 6B )= 6 

6B = 18 

x (t ) = e − 2t (1 + 5cos 6t + 3sin 6t )

[not required: graph of the complete solution]

B= 3

ENGI 3424 1 (a)

Mid Term Test - Solutions

Page 3 of 10

(continued)

Additional notes: In the method of undetermined coefficients, if the incorrect choice of trial particular solution is made, then no consistent value of the constant c can emerge: !

xP = ct e −2t

e −2t

 x

Substitute into

P

= 36 e−2t :

((−4 + 4t ) + 4 (1− 2t ) + 40t )c e −2t = 36e − 2t

 (36t + 0 )c = 0t + 36  36c = 0 and 0 = 36 which obviously has no solution for c. 1 It is incorrect to claim 36ct = 36  c = t because c is a constant, not a function of t. However, this second error cancels out the first error! 1 1 2t 2t c=  xP = t e− = e− which is the correct particular solution. t t

There is a formula for the particular solution of the ODE constants and r, k are non-zero constants): r xP ( t ) = 2 ekt k +k p +q

re

kt

(where p, q are

provided that k 2 + k p + q  0 , (which is equivalent to ek t not being a constant multiple of either component of the complementary function). It is not difficult to prove that this formula works. In question #1, p = 4, q = 40, r = 36, k = − 2 36 36 −2t 2t e−2t = e  xP ( t ) = = e− 4 − 8 + 40 36

ENGI 3424 2 (a)

Mid Term Test - Solutions

Page 4 of 10

Use the RK4 (Runge-Kutta) method with a single step (h = 0.5) to estimate the value of y at x = 0.5 to three decimal places, given that y ( 0 ) = 1 and that y is the solution of the ordinary differential equation dy = x2 + y dx dy Note: the RK4 algorithm for = f ( x, y) , y ( xo ) = yo includes dx k1 = f ( xn, yn )

( f ( xn + 12 h ,

[10]

) yn + 12 h k )

k2 = f xn + 12 h, yn + 12 h k1 k3 =

2

k4 = f ( xn + h , yn + h k3 ) h y n+1 = y n + (k 1 + 2k 2 + 2k 3 + k 4 ) 6 (b) Confirm your answer by solving the ordinary differential equation exactly.

(a)

[12]

xo = 0, yo = 1, h = 0.5  x1 = 0.5

k1 = f ( xo , yo ) = xo2 + yo = 0 +1 = 1.000000

( ) ( ) ( ) = f ( xo + 12 h, yo + 12 hk ) = ( xo + 12 h) + ( yo + 12 hk ) = ( 0.25) 2

k2 = f xo + 12 h, yo + 12 hk1 = xo + 12 h + yo + 12 hk1 = ( 0.25) + 1.25 = 1.312500

k3

2

2

2

2

2

+ 1.328125 = 1.390625

k4 = f ( xo + h, yo + hk3 ) = ( xo + h) + ( yo + hk3 ) = ( 0.5) +1.6953125 = 1.9453125 2

y1 = y o +

2

h 0.5 ( k 1 + 2k 2 + 2k 3 + k 4 ) = 1+ (1+ 2.625+ 2.78125+ 1.9453125 ) 6 6

0.5 ( 8.3515625 ) = 1 + 0.695963 6 correct to 3 d.p., y (0.5 ) = 1.696 = 1+

One must keep greater precision than three decimal places in all intermediate calculations, otherwise rounding errors can propagate and cause an error in the final answer.

ENGI 3424

2 (b)

Mid Term Test - Solutions

dy dy 2 2 = x + y  − y = x which is linear, but not separable. dx dx P = − 1  h =  P dx =  − 1dx = − x  e h = e− x

e

h R dx

(

=  e −x x2 dx

)

= − x 2 + 2x + 2 e− x

y = e−h

(  e hR dx + C )

((

)

= e x − x2 + 2 x + 2 e − x + C

)

 y ( x) = C e − x −2 x −2 x

2

y ( 0 ) = 1  1 = C − 0 − 0− 2  Complete solution: x y ( x) = 3 e − x 2 − 2 x − 2

C= 3

 y( 0.5) = 3 e0.5 − 0.25 − 1 − 2 = 1.696163 The RK4 estimate is identical to the third decimal place (but not the fourth).

OR The Chapter 2 method can be adapted to this first order ODE: dy − y = x2 dx A.E.:  − 1= 0   = 1 C.F.: y = Ae x P.S.: r = x 2 Therefore try y P = bx2 + cx + d  yP = 2bx+ c Substitute into y P − y P = x 2



(2bx + c )

− (bx 2 + cx + d ) = x 2 + 0 x + 0

Matching coefficients: x2 : − b = 1  b = −1

x1 : 2b − c = 0  c = 2b = − 2 x0 : c − d = 0  d = c = − 2  yP = − x2 − 2 x − 2 and the same general solution as before arises:

y ( x ) = Ae x − x2 − 2 x − 2

Page 5 of 10

ENGI 3424 3.

Mid Term Test - Solutions

Page 6 of 10

Find the equation of the family of orthogonal curves to the family of curves y = x 3+C and sketch two members of each family on the same diagram.

[5] [9]

dy = 3x 2 + 0 dx The orthogonal family must satisfy dy 1 1 1 x −1 = − 2 = − x−2  y = −  +A  dx 3x 3 3 −1 y = x 3 +C



y=

1 +A 3x

Many choices of A and C are possible, including those that produce this graph:

The important features of any sketch are: Every member of the original family (blue curves) has a point of inflexion with zero slope at its y axis intercept. Every member of the orthogonal family (red curves) has a vertical asymptote at the y axis and a horizontal asymptote at y = A. Every red curve intersects every blue curve (twice) at right angles. No two members of the same family ever meet each other.

ENGI 3424 4.

Mid Term Test - Solutions

Page 7 of 10

Find the complete solution to the initial value problem dy y = 1, y ( 0) = − 1 dx

[10]

The ODE is non-linear but is separable. dy y = 1  y dy = 1 dx   y dy =  1 dx  dx  y 2 = 2 x + A (where A = 2c)

y ( 0) = − 1  + 1= 0+ A 

y2 = x+ c 2

A= 1  y 2 = 2x + 1

Note: the explicit forms y =  2 x + 1 and y = − 2 x + 1 are also correct, but y = 2 x + 1 is not correct (inconsistent with y ( 0) = − 1 )

Note: There is no possibility of a singular solution for this non-linear ODE, as there is no division by any function of y during the separation of variables. dy = 1. Also note that y  k (where k is any constant) is not a solution to the ODE y dx

ENGI 3424 5.

Mid Term Test - Solutions

Page 8 of 10

BONUS QUESTION

[+6]

Use the transformation of variables x = e t to find the general solution to 2

x2

d y dy − 3x + 4y = x2 2 dx dx

From page 2.36 of the lecture notes: dx = et = x Let x = et  dt dy dy dt dy dx dy 1 dy =  =  = e− t = By the chain rule, dx dt dx dt dt dt x dt

 x

dy dy = dx dt

d 2 y d  dy  d  dy  dx dy  1 d  − t dy  1  − t d 2y e = =  =  = − e −t        e 2 2 dx dx  dx  dt  dx  dt x dt  dt  x dt dt  2 1  d 2 y dy  d 2 y dy 2 d y = 2 2 −   x = − x  dt dt  dx 2 dt 2 dt d 2y dy + bx + cy = r (x ) , 2 dx dx (where b and c are constants) therefore transforms into the ODE  d 2 y dy  dy d2 y dy b cy r x − + + =  + ( b− 1) + cy = r ( x ) ( )  dt 2 dt  2 dt dt dt  

Any Cauchy-Euler ODE of the type x2

Here b = −3, c = 4,

( )

r ( x ) = x2 = et

2

= e2 t

A.E.:

d 2y dy 2 − 4 + 4y = e t 2 dt dt 2  2 − 4  + 4 = 0  (  − 2 ) = 0   = + 2, + 2

C.F.:

yC (t ) = ( At + B )e 2 t

The equivalent ODE is

P.S. by the method of undetermined coefficients: 2t 2t r( t) = e but e2t and t e are both parts of the C.F. Therefore try

y P (t ) = c t 2e 2t 

)

+ 4t 2 e 2t

y

y P (t ) = c  c=

1  2

y P (t ) =

( ( 2+ 8t + 4t 2 )− 4 (2t+ 2t 2 )+ 4t 2 )e 2t =

1 2 2t t e 2

2c e

2t

= e

2t

ENGI 3424 5.

Mid Term Test - Solutions

Page 9 of 10

(continued)

OR P.S. by the method of variation of parameters:

y1 = e 2t , y 2 = t e 2t , r = e 2t

W= W1 =

u

y1 y

=

r

−t e4t

W W

4t

y1 y

t e2t

e 0

2t

= (1+ 2t − 2t )e 4t = e 4t

u= −

1 2 t 2

( )

= + y1r = e 2t e 2t = e 4 t

e 4t t

e4

= −t 

= 1 

y P = u  y1 + v  y2 = −

G.S.:

(1 + 2 t ) e

t

= − y r = − t e2 t e 2 t = − t e 4 t

y

W W

e2t

( )

0 y2

W2 =

v

y2

v= t

1 2 2t 1 t  e + t  t e 2t = t 2 e 2t 2 2

1 y ( t ) =  t 2 + At + B  e 2t 2 

but x = e t  t = ln x 

2 1  y ( x ) =  ( ln x ) + A ln x + B  x 2 2 

Note that alternative methods exist to solve these Cauchy-Euler ODEs, including one that begins with the substitution y = x r , but these methods are far beyond the scope of ENGI 3424 and are excluded by the wording of this question, which requires the use of the transformation x = e t .

ENGI 3424 5.

Mid Term Test - Solutions

Page 10 of 10

Additional note: The question does not require verification of this solution. However, such verification does provide good practice at the rules of differentiation, especially the product rule. 2 1  y ( x ) =  ( ln x ) + A ln x + B x 2 2   2 1 1 2   1   y ( x ) =  ( ln x ) + A + 0  x2 +  ( ln x ) + A ln x + B  ( 2 x ) x x 2 2     2 1   y ( x ) = x ( ln x + A) + 2x  ( ln x ) + A ln x + B  2   2 1  1   ln x A  y ( x) = ( ln x + A) + x  + 0  + 2  ( ln x ) + A ln x + B  + 2 x  +  x x  2   x 2 1   y ( x) = 1 + 3 ( ln x + A) + 2  ( ln x) + Aln x + B  2 

 x 2 y ( x ) − 3x y  ( x ) + 4y ( x ) = 2 1  x2 + 3x2 (ln x + A ) + 2 x2  ( ln x ) + A ln x + B  2 

1 2 − 3x 2 ( ln x + A ) − 6x2  ( ln x ) + A ln x + B  2   2 1  + 4x 2  ( ln x ) + A ln x + B  = x2 = r ( x) 2 

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