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Multiple-Choice Test
Euler’s Method
Ordinary Differential Equations
COMPLETE SOLUTION SET
- To solve the ordinary differential equation
3 5 sin , ( ) 0 5
2
dy
by Euler’s method, you need to rewrite the equation as
(A) sin 5 , ( ) 0 5
2 = x − y y = dx
dy
(B) (sin 5 ), ( ) 0 5 3
= x − y y = dx
dy
(C) , ( ) 0 5 3
cos 3
3
^ =
= − − y
y x dx
dy
(D) sin , ( ) 0 5 3
= x y = dx
dy
Solution
The correct answer is (B).
To solve ordinary differential equations by Euler’s method, you need to rewrite the equation in
the following form
f ( x , y ), y ( ) 0 y 0 dx
dy = =
Thus,
3 5 sin ,^ ( ) 0 5
2
dy
( )
(sin 5 ), ( ) 0 5 3
3 sin 5 , 0 5
2
2
x y y dx
dy
x y y dx
dy
- Given
3 5 sin , ( 0. 3 ) 5
2
dy
and using a step size of h = 0. 3 , the value of y ( 0. 9 )using Euler’s method is most nearly
(A) − 35. 318
(B) − 36. 458
(C) − 658. 91
(D) − 669. 05
Solution
The correct answer is (A).
First rewrite the differential equation in the proper form.
2
2
sin 5 3
sin 5 3
f x y x y
x y dx
dy
Euler’s method is given by
y i + 1 = yi + f^ (^ xi , yi ) h
where
h = 0. 3
For i = 0 , x (^) 0 = 0. 3 , y 0 = 5
sin 0. 3 55 0. 3 3
2
1 0 0 0
= + − ×
= + ×
f
y y f x y h
y 1 is the approximate value of y at
x = x 1 = x 0 + h = 0. 3 + 0. 3 = 0. 6
For i = 1 , x 1 (^) = 0. 6 , y 1 =− 7. 4704
sin 0. 6 5 7. 4704 0. 3 3
2
2 1 1 1
=− + − − ×
=− + − ×
f
y y f x y h
y (^) 2 is the approximate value of y at
x = x 2 = x 1 + h = 0. 6 + 0. 3 = 0. 9
y ( 0. 9 ) ≈− 35. 318
y ( 0. 9 ) ≈ 4. 7647
Thus
- 9
1
× e dx
dy
e y dx
dy (^) x
- The velocity (m/s) of a body is given as a function of time (seconds) by
v ( ) t = 200 ln( 1 + t ) − t , t ≥ 0
Using Euler’s method with a step size of 5 seconds, the distance traveled in meters by the body
from t = 2 to t = 12 seconds is most nearly
(A) 3133. (B) 3939.
(C) 5638.
(D) 39397
Solution
The correct answer is (A).
( ) ( )
( )
f ( tS ) ( t ) t
t t dt
dS
vt t t
, 200 ln 1
200 ln 1
200 ln 1
Euler’s method is given by
S (^) i + 1 = Si + f ( t (^) i , Si ) h
where
h = 0. 5
For i = 0 , t 0 (^) = 2 s, S 0 (^) = 0 m(assuming S (^) 0 = 0 mwould make S (^) 2 the value of the distance
covered, as the distance covered is S (^) 2 − S 0 )
( )
( )
( ( ) )
- 6 m
0 200 ln 1 2 2 5
1 0
1 0 0 0
= + + − ×
= + ×
= + ×
t t h
f
S S f t S h
For i = 1 , t 1 (^) = 7 s, S 1 = 1088. 61 m
( )
( )
( ( ) )
- 1 m
- 6 200 ln 1 7 7 5
2 1 1 ,^1
= + + − ×
= + ×
= + ×
f
S S f t S h
- 1 m
S − S ≈ S − S
- Euler’s method can be derived by using the first two terms of the Taylor series of writing the
value of yi (^) + 1 , that is the value of y at xi (^) + 1 , in terms of yi and all the derivatives of y at xi. If
h = xi + 1 − x i , the explicit expression for yi (^) + 1 if the first three terms of the Taylor series are
chosen for the ordinary differential equation
2 3 , ( ) 0 7
5
− y e y dx
dy (^) x
would be
(A) y y ( e yi ) h
x i i
i (^) 3 2
1 =^ + −
−
(B) ( )
5 5 2 1 2
y y e i^ y h e i h
x i
x i i
− −
(C) ( )
5 5 2 1 4
y y e y h e yi h
x i
x i i
i i
− −
(D) ( )
5 2 1 2
y y e yih yih
x i i = + − i^ − −
Solution
The correct answer is (C).
The differential equation
2 3 , ( ) 0 7
5
− y e y dx
dy (^) x
is rewritten as
( ) ( )
f ( x y ) ( e y )
e y y dx
dy
x
x
5
5
−
−
The Taylor series is given by
( ) ( ) ( ) ... 3!
1 ,
3
3 2 1 ,
2
2
1 ,
- 1 =^ + + − + + − + i + − i + xy
i i x y
i i x y
i i x x dx
d y x x dx
d y x x dx
dy y y
i i i i i i
( ) ( ) ''( , )( ) ... 3!
3 1
2 yi + 1 = yi + f xi yi xi + 1 − xi + f xi yi xi + 1 − xi + f xi yi xi + − xi +
If we look at the first three terms of the Taylor series
( ) ( )
( ) ( )
2
2 1 1 1
y f x y h f x y h
y y f x y x x f x y x x
i i i i i
i i i i i i i i i i
+ =^ + + − + +−
where
h = xi + 1 − x i
( )
( ) ( )
e y
e e y
dx
dy
y
f
x
f f x y
x
x x
5
5 5
−
− −
then the value of yi (^) + 1 is given by
( )
5 5 2 1 4
y y e y h e yi h
x i
x i i
i i
− −
( )
( )
( ) 98. 6
θ B
where
B = time of death
we get
85 72
6 = +
− k × Ae (1)
78 72
9 = +
− k × Ae (2)
- 6 = + 72
− k × B Ae (3)
Use Equations (1) and (2) to find A and k.
6
6
−×
−×
k
k
Ae
Ae (4)
9
9
−×
−×
k
k
Ae
Ae (5)
Dividing Equation (4) by Equation (5) gives
( ( ))
hours
ln 2. 1667 3
3
9
6
−
k
e
Ae
Ae
k
k
k
Knowing the value of k , from Equation (5)
A = 61. 028 °F
Substitute k and A into Equation (3) to find B.
( )
- 2220 hours
ln 26. 6 ln 61. 028 0. 25773
25773
25773
− ×
− ×
−×
B
B
B
e
e
Ae
B
B
kB
Note to the student:
You can also do the problem by assuming that the initial time reference is zero, and that the
temperature then is θ ( 0 )= 98. 6. Then the temperature is given at the time the body was found
as θ ( C )= 85 °F, and that θ ( C + 3 )= 78 °F. You can now find k, A and C just like as given
above_._ The value of C in fact is the time between the body was found and the time of death_._
You will get C = 2. 7780 hrs_._
The time of death is 3.2220 hrs from 12 noon, that is 3 :( 0. 2220 × 60 )PM=3:13 PM.
