Review: Chapters 9 & 17
Monday, May 4
9.1-9.2: Differential Equations and Direction Fields
1. Show that the function y = x sin x satisfies the differential equation y 00 + y = 2 cos x.
y 00 + y = (x sin x)00 − x sin x
= (sin x + x cos x)0 + x sin x
= cos x + cos x − x sin x + x sin x
= 2 cos x
2. Show that the function y = xe−x + 2 satisfies the differential equation y − xy 0 = x2 e−x + 2.
y − xy 0 = xe−x + 2 − x(xe−x + 2)0
= xe−x + 2 − x(e−x − xe−x )
= xe−x + 2 − xe−x + x2 e−x
= x2 e−x + 2
3. If y 0 = y and y(0) = 1, estimate y(1) using Euler’s method with a step size of ∆x = 0.5.
y(0) = 1
y 0 (0) = 1
y(0.5) ≈ y(0) + ∆x · y 0 (0)
y(0.5) ≈ 1.5
y 0 (0.5) = 1.5
y(1) ≈ y(0.5) + ∆x · y 0 (0.5)
y(1) ≈ 1.5 + 0.5 · 1.5
y(1) ≈ 2.25
The solution to the given differential equation is y = ex , and so y(1) = e ≈ 2.718. To get a better
approximation from Euler’s method we would have to use a much smaller step size.
4. Match the following differential equations with the appropriate direction fields:
(a) y 0 = x + y − 1
(c) y 0 = x2 + y 2 − 1
(e) y 0 = x(y − 1)
(b) y 0 = xy − 1
(d) y 0 = x − y + 1
(f) y 0 = x2 − y 2 + 1
1
1. C
4. B
2. E
5. A
3. F
6. D
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9.3: Separable Equations
1. y 0 = x/y, y(0) = −3
dy
= x/y
dx
y dy = x dx
Z
Z
y dy = x dx
1 2
1
y = x2 + C
2
2p
y = ± x2 + C
p
y = − x2 + 9
2. y 0 = xy sin x, y(0) = 1
dy
= xy sin x
dx
1
dy = x sin x dx
y
Z
Z
1
dy = x sin x dx
y
ln y = −x cos x + sin x + C
y = e−x cos x+sin x+C
y = e−x cos x+sin x
3. xy 0 − y = 1, y(2) = 3
dy
=1+y
dx
dy
dx
=
1+y
x
ln(1 + y) = ln x + C
x
1 + y = eln x+C
y = Kx − 1
y = 2x − 1
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9.5: Linear Equations
1. xy 0 − y = 1, y(2) = 3
y0 −
1
1
y=
x
xR
I = e −1/x
= e− ln x
= 1/x
Z
y = ( IQ)/I
Z
= x( 1/x2 )
= x(−1/x + C)
= Cx − 1
y = 2x − 1
Unsurprisingly, this is the same answer as for the previous problem.
2. 2xy 0 + y = 6x, y(4) = 20
y0 +
1
y=3
2x
R
I = e 1/2x
1
= e 2 ln x
√
= x
Z
y = ( IQ)/I
Z
√
1
= √ ( 3 x)
x
1
= √ (C + 2x3/2 )
x
√
= 2x + C/ x
√
y = 2x + 24/ x
3. y 0 + xy = x, y(1) = 1.
I=e
R
x
2
= ex /2
Z
y = ( IQ)/I
Z
2
2
= e−x /2 ( xex /2 )
= e−x
2
/2
(ex
2
/2
−x2 /2
= 1 + Ce
y=1
4
+ C)
17.1-2: Second-order Linear Equations
Solve each non-homogeneous equation using either variation of parameters or the method of undetermined
coefficients, whichever is more appropriate.
1. y 00 + 3y 0 + 2y = sin x + 2 cos x
Undetermined coefficients:
y = A sin x + B cos x
y 0 = A cos x − B sin x
y 00 = −A sin x − B cos x
y 00 + 3y 0 + 2y = (A − 3B) sin x + (3A + B) cos x
(A − 3B) = 1
3A + B = 2
A = 7/10
B = −1/10
7
1
yp =
sin x −
cos x
10
10
yc = C1 e−x + C2 e−2x
1
7
sin x −
cos x + C1 e−x + C2 e−2x
y=
10
10
2. y 00 − 2y 0 + y = 2xex − ex
The general solution to y 00 − 2y 0 + y = 0 is yc = C1 ex + C2 xex , so we’ll have to try multiplying by x
or x2 :
y = Ax3 ex + Bx2 ex
y 0 = Ax3 ex + (3A + B)x2 ex + 2Bxex
y 00 = Ax3 ex + (6A + B)x2 ex + (6A + 4B)xex + 2Bex
y 00 − 2y 0 + y = 6Axex + 2Bex
A = 1/3
B = −1/2
1
1
y = x3 ex − x2 ex + C1 ex + C2 xex
3
2
3. y 00 + y = 2 sin x + 3
The general solution to y 00 + y = 0 is y = A sin x + B cos x, so try multiplying by x to solve for
y 00 + y = 2 sin x. Separately, try y = Cx2 + Dx + E to solve y 00 + y = 3.
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y = Ax sin x + Bx cos x + Cx2 + Dx + E
y 0 = A sin x + Ax cos x + B cos x − Bx sin x + 2Cx + D
y 00 = 2A cos x − Ax sin x − 2B sin x − Bx cos x + 2C
y 00 + y = −2B sin x + 2A cos x + Cx2 + Dx + (2C + E)
B = −1
A=0
C=0
D=0
E=3
yp = −x cos x + 3
y = −x cos x + 3 + A sin x + B cos x
4. y 00 + y =
1
cos x
y1 , y2 , = cos x, sin x
W =1
Z
Z
1
sin x + sin x 1
cos x
= − cos x(− ln cos x + C1 ) + sin x(x + C2 )
y = − cos x
= x sin x + cos x ln cos x + C1 cos x + C2 sin x
5. y 00 + 3y 0 + 2y = sin(ex )
y1 , y2 , = e−x , e−2x
W = −e−3x
Z
Z
−x
x
x
−2x
y = −e
−e sin e + e
−e2x sin ex
= e−x (− cos ex + C1 ) + e−2x (ex cos ex − sin ex + C2 )
= −e−2x sin ex + C1 e−x + C2 e−2x
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