page 1 of Section 4.1 solutions
SOLUTIONS Section 4.1
1. (a) y' +
4
-6
y =
,
x+2
(x+2)2
n(x+2)4
I = e
= (x+2)4,
answer is y =
4
,
x+2
£P(x) dx = 4 n(x+2) = n(x+2)4
-6(x+2)3
(x+2)4y = £ -6(x+2)2 dx =
3
-2
k
+
x+2
(x+2)4
(b) y' + 4xy = x,
P(x) =
P(x) = 4x,
£P(x) dx = 2x2,
+ k,
2
I = e2x ,
2
2
1
2
1
2
ye2x = £ xe2x dx = 4 e2x + K. Answer is y = 4 + Ke-2x
1
(c) method 1 Rewrite the equation as y' + y = 2 e2x. Then
1
1
1
P(x) = 1, I = ex, yex = £ex 2 e2x dx= 2 £e3x dx = 6 e3x + K.
Answer is y =
1 2x
e
+ Ke-x.
6
(since the coeffs are constant) (no need to rewrite the equation)
m = -1, yh = Ae-x.
1
Try yp = Be2x. Then 4Be2x + 2Be2x = e2x, 6B = 1, B = 6
method 2
yp =
1 2x
e
, answer is y =
6
(d) y' - y cot x = csc x,
I = en csc x = csc x,
Answer is y = - cos x
Ae-x +
1 2x
e
6
P(x) = - cot x, £ P(x) dx = -n sin x = n csc x,
y csc x = £ csc2x dx = - cot x + K
+ K sin x
1
y' = 4. Then £P(x) dx = n x,
x
K
I = en x = x,
xy' = £ 4x dx = 2x2 + K so y' = 2x + . Antidiff to get
x
y = x2 + K n x + C (two arbitrary constants)
2. Rewrite as x(y')' + y' = 4x,
(y')' +
3. (a) y' + 4xy = x
P(x) = 4x, Q(x) = x
” P(x) dx = x2 + K
’
2
2
2
2
I = ex +K = ex eK which you could call Aex but I'll just leave it ex eK
(doesn't make any difference)
Iy = ”’ IQ
2
2
x2 x dx
ex eK y = ”’ ex eK x dx = eK ”e
’
2
2
The eK on both sides cancels out now and you're left with ex y = ”’ ex x dx
etc.
Any K you put in at the £P(x) dx stage will cancel out later so why bother. The
only constant you'll have left at the end of the problem is the one you put in
when you find ”’ IQ.
page 2 of Section 4.1 solutions
(b) You get solution y = 1/4. This is a solution but the general solution. If you
have an IC to satisfy, unless you are very very lucky, your one particular
solution won't satisfy that IC and you have no constants available to make it
satisfy the IC.
4. method 1 y' - ky = 0,
method 2
y' - ky = 0,
m = k,
y = Aekx
P(x) = -k,
ye-kx = £ 0 dx = C,
5. method 1
£ P(x) dx = -kx,
I = e-kx,
y = Cekx
P(x) = 1, £ P(x) = x,
I = ex,
yex
= £ 1 dx = x + K,
y = xe-x + Ke-x
3 + e
e
xe-x + 3e-x-1
To get y(-1) = 3 need 3 = -e + Ke , K =
3 + e
e -x =
+ e-x
e
method 2 (since coeffs are constant) m = -1, yh = Ae-x.
Try yp = Bxe-x (step up). Need
-Bxe-x + Be-x + Bxe-x = e-x
xe-x terms drop out.
Equate coeffs of the e-x terms
B = 1
Answer is
y =
xe-x +
Gen sol is y = Ae-x + xe-x .
The IC determine A as in method 1.
6. (a) eK is not quite arbitrary. It can never be zero and it can never be
negative. So it really shouldn't be turned into a C which is totally arbitrary.
But most people pay no attention to these niceties. And it usually works out OK
in the long run. You would probably find that the new not-so-arbitrary constant
C (that used to be eK ) comes out positive anyway when you plug in a realistic
condition.
(b) Yes because n K can take on any value, from very negative to very
positive.
On the other hand, you have a slight problem because you shouldn't be taking log
of an arbitrary K because you can't take n of a negative number or zero. Most
people don't worry about this either.
7. y' +
x2 + 1
2
2
y =
. Then P(x) = ,
x
x
x
£ P(x) = 2 n x = n x2,
x4
x2
x2y = £x(x2 + 1) dx = £(x3 + x) dx =
+
4
2
+ K,
y =
I = x2,
x2
1
K
+
+
4
2
x2
1
1
y = x.
P(x) = - , £ P(x) = -n x = n x-1, I =
x
x
1
y = £ dx = x + C, y = x2 + Cx.
x
1
1
Then solve y' y = 0 getting
y = £ 0 dx = K, y = Kx. So
x
x
8. First solve y' -
1
,
x
page 3 of Section 4.1 solutions

y = 

x2 + Cx
Kx
if x ≤ 3
if x > 3
The condition y(1) = 2 makes C = 1.
To get the sol continuous, we want x2 + x = Kx when x = 3,
 2
Answer is y =  x + x
 4x
if x < 3
if x > 3
12 = 3K, K = 4
page 4 of Section 4.1 solutions
HONORS
9.(a) The equation is y' - ry = -h where r and h are (positive) constants. It's
a linear first order DE with constant coeffs.
method 1 for solving the DE
m = r, yh = Aert
Try yp = B. Need 0 - rB = -h, B = h/r
h
ygen = Aert +
r
method 2 for solving the DE
P = -r, Q = -h, I = e-rt
e-rt y = ”’ -h e-rt dt =
y =
h -rt
e
+ A
r
h
+ Aert
r
Whichever method you used, plug in the IC y(0) = N to get A = N Solution is y(t) =
h
h
+ (N - )ert
r
r
h
r
1
1
(b) The solution now is y(t) = 2 h + (40 - 2 h)e2t
Fished out means that as t gets larger, y hits 0 eventually. So the problem is
to find which h's let y(t) reach 0.
1
1
Here's the graph point of view. The graph of y(t) = 2 h + (40 - 2 h)e2t starts
at height 40 when t = 0. It's decreasing if
1
40 - 2 h < 0
h > 80
and it's increasing if h < 80.
So the lake gets fished out if h > 80.
To illustrate, here's the graph of y(t) for h = 60 (and you can see the fish
population taking off) and again for h = 90 where you see the population hit 0
(at which point the mathematical model ceases to apply.)
page 5 of Section 4.1 solutions
Plot[60/2 + (40 - 60/2)E^(2t),{t,0,2}];
40
1
2
Plot[100/2 + (40 - 100/2)E^(2t),{t,0,1.5}, Ticks-{{.5,1,1.5},{20,40}}];
40
0.5
1
1.5
100
100 2t
1
+ (40 )e
= 0, e2t = 5, t = 2n 5
2
2
In the second diagram, the graph crosses the horizontal axis at t =
That's when the lake is fished out.
(c)
1
n 5.
2
page 1 of Section 4.2 solutions
SOLUTIONS Section 4.2
1
sin y = - 2 x2 + C
1. (a) cos y dy = -x dx,
(b)
y dy = -
dx
,
x3
(c) y4 dy = -x2 dx,
dy
y
(d)
=
dx
,
2x + 3
Ky = √
2x + 3 ,
1 2
1
y =
2
2x2
+ C,
√
1
+ D
x2
y = –
1 5
1
y = - 3 x3 + C, y =
5
(implicit sol)
5
5
K - 3 x3
√
1
= n √
2x + 3 ,
n Ky = 2 n(2x + 3)
y = A√
2x + 3
warning It's OK to write
1
n y = 2 n(2x + 3) + C
but when you take exp on both sides it is wrong to get
y
=
√
2x + 3
PLUS eC
WRONG
which turns into
y =
√
2x + 3 + A
The right way to take exp is to get
y =
which turns into
y =
dx
e-y dy =
,
x2
1
y = - n( + D)
x
(e)
n(2x+3)1/2 + C
e
n(2x+3)1/2
e
- e-y
(f) y dy = (5x + 3) dx,
= -
1 2
y =
2
RIGHT
TIMES
1
+ C,
x
C
e
=
A√
2x + 3
1
e-y =
+ D,
x
5 2
x + 3x + C,
2
y = –
1
-y = n( + D),
x
√
5x2 + 6x +
(g) not separable
(h) (y + 1) =
y =
1
dx,
x
1 2
y + y = n Kx
2
(implicit sol),
-2 – 
√
4 + 8 nKx
= -1 – 
√
1 = 2  Kx
2
(explicit sol)
D
page 2 of Section 4.2 solutions
1
2. For (e), y = - n ( + D)
x
x2 dy = x2 y'(x) dx = x2…
1
-n ( + D)
x
e-y dx = e
-1
1
dx
… dx =
1
2
1
x
+ D
+ D
x
x
1
n ( + D)-1
x
= e
1
dx = ( + D)-1 dx
x
So x2 dy does equal e-y dx, QED.
For (f), y = –
√
5x2 + 6x +
so y' does equal
5x + 3
,
y
D , y' =
10x + 6
5x + 3
=
–2√

5x2 + 6x + D
–√

5x2 + 6x + D
QED.
x2/2
x2/2
dy
1
= x dx, n Ky = 2 x2,
Ky = e
,
y = Ae
y
Use the condition to get 3 = Ae1/2 , A = 3e-1/2 ,
-1/2 x2
(x2-1)/2
Sol is y = 3e
e ,
y = 3e
1 2
5
(b) y dy = (3 - 5x)dx,
y = 3x - 2 x2 + C. Set x = 2, y = 4 to get C = 12.
2
3.(a)
1 2
5
Then
y = 3x - 2 x2 + 12,
y = 
√
6x - 5x2 + 24
2
(Choose the positive square root since y is positive when x = 2.)
3
ey = 2 x2 + C.
Set x = 0, y = 2 to get C = e2.
3
3
Then ey = 2 x2 + e2,
y = n (2 x2 + e2)
dy
1
1
(d)
= cos x dx,
= sin x + C. Set x = 0, y = 2 to get C = - 24 .
y4
3y3
-1
Sol is y
3
1
3 sin x - 8
(c) ey dy = 3x dx,

√
4. method 12 (works because the coeffs are constant)
1
1
w' + 5 w = 2, m = - 5, wh = Ae-t/5
1
Try wp = B. Substitute into the DE to get 0 + 5 B = 2, B = 10
wgen = Ae-t/5 + 10 (which is the same as the answer 10 - De-t/5 from example 1
because the arbitrary constant A is the same as the arbitrary constant -D)
£P dt
1
method 2 P = 5 , Q = 2, I = e
= et/5
t/5 dt,
Iw = ”IQ
= ”2e
’
’
et/5 w = 10et/5 + K, w = 10 + Ke-t/5
5. The equation can be written as xy2 - y - 7 = 0 and treated as a quadratic
equation of the form ay2 + by + c = 0 where a = x, b = -1, c = -7. So
1 – 
√
1+28x
y =
, not really one implicit solution, but two. (And they real only if
2x
x > -1/28 and they aren't defined for x = 0.)
page 3 of Section 4.2 solutions
dy
= -10 dx, n Ky = -10x, Ky = e-10x, y = Ae-10x.
y
Plug in the IC y(0) = G to get y = Ge-10x.
1
Now let y = G/2 and find x: G/2 = Ge-10x, 2 = e-10x, -10x = n 1/2,
6.(a)
1
x = 10 n 2.
1
So the half life is 10 n 2.
(b) Let the constant of proportionality be called C. As in part (a), the half life
1
1
is
n 2.If you want this to be 3, choose C = 3 n 2.
C
7. (a) Differentiate w.r.t. x to get the differential equation of the family xy = K.
xy' + y = 0
y' = -y/x
The orthog family has differential equation y' = x/y. Solve it to find the
orthogonal family.
y dy = x dx
1 2
1
y = 2 x2 + A
2
y2 - x2 = C
Both families are hyperbolas (each hyperbola has two branches). In the diagram, the
original family is in darker type.
(b)
y
= A
x2
x-2 y' - 2x-3 y = 0
2y
y' =
x
The orthog family has DE y' = -
x
. Solve to get the orthog family.
2y
2yy' = -x
2y dy = -x dx
1
y2 = - 2 x2 + C
x2 + 2y2 = D (a family of ellipses)
K=2
C=-1/2
K=-1
A=3
A=1/2
C=1
C=3
9
D= 5
D=
D=1
A=-4
Problem 7(a)
Problem 7(b)
page 1 of Section 4.3 solutions
SOLUTIONS Section 4.3
y
1. d( ) comes out to be (22) immediately by the quotient rule and similarly for (23)
x
d(x2 + y2)-1 = -(x2 + x2)-2 d(x2 + y2))
(chain rule)
= -(x2 + x2)-2
(2x dx + 2y dy) which is (24)
1
d(–√

x2 + x2 ) = – 2 (x2 + y2)-1/2 d(x2 + y2)
(chain rule)
2x dx + 2y dy
=
which cancels to (25)
2
2
–2 √

x + x
d(n (x2 + y2) =
d(arctan y/x) =
2.
1
2
x + y2
d(x2 + y2)
1
y
d( ) =
2
x
1 + (y/x)
(chain rule) =
1
1 + (y/x)2
x dy
2x dx + 2y dy
(x2 + y2)
- y dx
x2
=
-y dx + x dy
x2 + y2
x = r cos œ so by (1), dx = cos œ dr - r sin œ dœ
y = r sin œ, dy = sin œ dr + r cos œ dœ
ºp
= 2x. Not equal so form is not exact
ºy
1
Exact. Antidiff p w.r.t. x to get 4 x4 + x3y. Diff this
3.(a) p = 2xy, q = y,
ºq
ºx
= 0,
ºq
ºp
=
= 3x2.
ºx
ºy
temporary answer w.r.t. y to get x3. Compare with q and see that you should tack on
1 4
y . Answer is
4
(b)
f(x,y)
(c) f(x,y) = -
y
+ 5y
x
1
1
= 4 x4 + x3y + 4 y4
ºq
ºp
ºq
3
=
,
= 3xy2, q = 2 x2y2 + any f(y)
ºx
ºy ºx
3
For example q could be 2 x2y2 + sin y + 7
4. Need
5. (a) d(2x3 + xy2 + y3) = 0, implicit sol is
2x3 + xy2 + y3 = C
C - x3
Implicit sol is x3 + xy = C. Explicit sol is y =
x
(x-y cos x) dx - (y + sin x) dy = 0
(b) d(x3 + xy) = 0
(c)
1
1
d(2 x2 - y sin x - 2 y2) = 0
1
1
Implicit solution is 2 x2 - y sin x - 2 y2 = K
(d) exy dx - dy= 0
ºq
ºp
= 0 but
= xexy
ºx
ºy
r2 sin œ dœ = 0
0
cos œ - r = K
Not exact since
(e) (2r cos œ - 1)dr d(r2 cos œ - r) =
Implicit sol is r2
(f) not exact
1
(g) d(sin x cos y) = d(4 x4)
1
Implicit sol is sin x cos y = 4 x4 + C
page 2 of Section 4.3 solutions
(h) (ye-x - sin x) dx - (e-x + 2y) dy = 0
d(-ye-x + cos x - y2) = 0
Implicit sol is -ye-x + cos x - y2 = C
6.
-ye-x + cos x - y2 = C to get
Take differentials throughout
-y…e-x dx + e-x…-dy - sin x dx - 2y dy = 0
(ye-x - sin x) dx =
Collect terms:
(e-x + 2y) dy, QED
1
7.(a) d(x2y + 2 y2) = 0
1
Implicit sol is x2y + 2 y2 = C
Set x = 1, y = 4 to get C = 12.
(b)
d(-cos(2x + 3y) ) = 0
Implicit particular sol is is
Implicit sol is -cos(2x + 3y) = C
Set x = 0, y = π/2 to get C = 0.
(c)
d n(x + y) = d(x).
Set x = 0, y = 1
1
x2y + 2 y2 = 12
Implicit particular sol is cos(2x + 3y) = 0
Implicit sol is
n(x + y) = x + C.
to get C = 0. Implicit particular sol is
n(x + y) = x.
Then x + y = ex so explicit solution is y = ex - x
1
3
8.(a) d(3 x3 + 2x + 2 y2) = 0
(b) (x2 + 2)dx = -3y dy,
1 3
3
x + 2x + 2 y2 = K
3
Implicit sol is
1 3
3
x + 2x = - 2 y2 + K
3
9. (a) It doesn't do any good to collect the dx terms and get
(x2 + y2 + y) dx - x dy = 0 because this arrangement isn't exact.
1
Instead, look at (27). Use integrating factor
. The equation becomes
2
x + y2
x dy - y dx
y
dx =
so x = tan-1
+ K
x
x2 + y2
(b)
See (22).
Use integrating factor 1/y2.
y dx - x dy
= dx,
y2
(c) See (25).
dy =
x
=
y
x + K,
Use integrating factor
x dx + y dy
,

√
x2 + y2
y =
√
x2 + y2
y =
1

√
x2 + y2
+ K
(d) x dx = (x2 + y2 - y) dy
x dx + y dy = (x2 + y2)dy
See (26). Use integrating factor
2
.
2
x + y2
x
x + K
.
page 3 of Section 4.3 solutions
2x dx + 2y dy
x2 + y2
= 2 dy,
n(x2 + y2) = 2y + K
(e) See (23). Multiply by 1/x2 to get
d(
x dy - y dx
= 2x dx + 2y dy,
x2
y
y
) = d(x2 + y2),
= x2 + y2 + K. Implicit sol is y =
x
x
x3 + xy2 + Kx.
page 1 of Section 4.4 sols
SOLUTIONS Section 4.4
1. Remember that at each point (x,y), the idea is to draw a little segment with
slope x/y.
On the line y = x, all the segments have slope 1.
In quadrants I and III the segments all have positive slope.
In quadrants II and IV the segments have negative slope
As you move right on a horizontal line in quadrants I, y stays the same and x gets
larger so the segments get steeper. As you move left on a horizontal line in
quadrant II, y stays the same and x gets negatively larger so the segments get
steeper (but with negative slope).
As you move up a vertical line in quadrant I, x stays the same and y gets larger
so the segments get less steep. etc.
MyField = directionField[x/y, {x,-3,3},{y,-3,3},.5,.3];
Show[MyField, Axes->True, Ticks->None];
y dy = x dx
y2
x2
=
+ K
2
2
y2 - x2 = A
Each solution is a hyperbola. Here's a picture of the direction field along with
the two particular solutions y2 - x2 = 2, y2 - x2 = -2
SomeSols = ImplicitPlot[{y^2 - x^2== -2,y^2 - x^2 ==2},{x,-3,3},
PlotStyle->{{GrayLevel[.5],Thickness[.01]}},
PlotRange->{-3,3},DisplayFunction->Identity];
Show[{MyField,SomeSols}, Axes->True,
Ticks->None,DisplayFunction->$DisplayFunction];
page 2 of Section 4.4 sols
(b) When y = x, the little segments have slope 1.
As you move right on a horizontal line in quadrant I, y stays fixed and x
increases so the segments get less steep.
As you move up a vertical line in quadrant I, x stays fixed and y increases so
the segments get steeper.
dy
dx
=
y
x
n Ky = n x
Ky = x
y = Ax
The solutions are all lines through the origin.
2. Draw a curve through the point (0,1) using the little segments.
(By the way, the direction field in the problem happens to be that of the
DE y' = xy.)
3.(a). The segment at point (x0, y0) is supposed to have slope x0y0. If you solved
the DE and found the particular solution satisfying the condition y(x0) = y0, in the
vicinity of point (x0, y0) it's graph should look like the little segment.
(b) The equation is separable and linear first order and exact. I'll separate.
dy
= x dx
y
1
n Ky = 2 x2
2
Ky = ex /2
2
y = Aex /2
page 3 of Section 4.4 sols
2
2
2
(c) I plotted 2ex /2, 4ex /2 and -4ex /2.
A=4
A=2
A=-4
(d) Plug in the condition y(4) = 3; i.e., set x=4, y=3 to determine A.
-8+x2/2
2
3 = Ae8, A = 3e-8. The curve is y = 3e-8 ex /2; equivalently y = 3e
.
page 1 of solutions to review problems for Chapter 4
SOLUTIONS review problems for Chapter 4
1.(a) method 1 (exact)
y = –
1
3 2
1
3 2
d(3 x3 + 2x +
y )= 0, 3 x3 + 2x +
y = K
2
2
2
2
4
K - 9 x3 - 3 x
3
√

1
3
(x2 + 2) dx = -3y dy, 3 x3 + 2x = - 2 y2 + K ,
same y as in method 1 now.
dy
(b) method 1 (separable)
= - dx, n Ky = -x,
Ky = e-x, y = Ae-x
y
1
method 2 (exact) y dx + dy = 0 is NOT exact but dx +
dy = 0 is exact.
y
Then d(x + n y) = 0,
x + n y = K , n y = K - x,
y = eK-x ,
method 2 (separable)
y = eK e-x ,
y = Ae-x
method 3
(linear, constant coeffs)
method 4
(linear first order)
ex y = £ 0 dx = K,
y' + y = 0, m = -1, y = Ae-x
y' + y = 0,
P = 1, Q = 0, I = e£P = ex,
y = Ke-x
(c) method 1 (can be arranged to be exact)
(y-2x) dx + x dy = 0,
d(xy - x2) = 0, xy - x2 = C
1
If x = 1, y = 2 then C = 1. Sol is xy - x2 = 1. Explicit sol is y = x +
x
1
method 2 (first order linear) y' +
y = 2, I = e£(1/x) = en x = x,
x
K
1
xy = £ 2x dx = x2 + K, y = x +
. If y(1) = 2 then K = 1. Sol is y = x +
x
x
(d) (linear)
y'' - y = 0, m = –1,
(e)
y'' - 3y' =
method 1
Try yp = Cx (step up)
y = Aex + Be-x
12, m = 0,3,
yh = A + Be3x
Get C = -4. Answer is y = A + Be3x - 4x
method 2 If you think of y' as the variable this is first order.
Let y' = u Then DE is u' = 3u + 12 This is exact, also separable,. also linear
with constant coeffs (P,Q stuff). Here's the separation method:
du
= dx,
3u + 12
1
n K(3u + 12) = x,
3
n K(3u + 12) = 3x,
K(3u + 12) = e3x
u = Be3x - 4
So y' = Be3x - 4 and (antidiff)
1
y = 3 Be3x - 4x + C ,
y = De3x - 4x + C, same as in other method
page 2 of solutions to review problems for Chapter 4
dy
= ex ey,
dx
(f) method 1 (separable)
e-y = A - ex,
e-y dy = ex dx,
-y = n(A - ex),
-e-y = ex + K,
y = - n(A - ex)
(exact) ex dx - e-y dy = 0,
y = - n(K - ex)
d(ex + e-y) = 0,
method 2
ex + e-y
= K,
(g) (Second order, variable coeffs, can't use m's)
method 1 Consider y' the variable (call it w if you like). Then the equation
is first order separable:
dw
dx
xw' - w = 1,
=
, n(w + 1) = n Kx, w + 1 = Kx, w = Kx - 1
w+1
x
So y' = Kx - 1,
y = Kx2 - x + C
The IC make K = 2, C = 1 Answer is y= 2x2 - x + 1
method 2 Consider y' the variable (call it w if you like). Then the equation
is first order linear:
1
1
w' w =
x
x
£P(x) dx
-n x
n 1/x
I = e
= e
= e
=
1
1 1
1
w = £
dx = + K.
x
x x
x
So
1
x
1
y = -x + 2 Kx2 + C .
y' = -1 + Kx,
The IC make
K = 4, C = 1,
Answer is y = -x + 2x2 + 1
2. method 1 (separable)
dv
= mg - cv
dt
dv
dt
= vc - mg
m
m
1
t
n(cv - mg) = c
m
K(cv - mg) = e-ct/m
cv - mg = A e-ct/m
mg
A -ct/m
+
e
c
c
c
Use the P,Q method. v' +
v = g,
m
v =
method 2
gm ct/m
ect/m y = £ gect/m dt =
e
c
method 3 (harder to notice)
m
cv - mg
p
dv
+ K,
y =
£ c/m ex
I = e
= ect/m,
gm
+ Ke-ct/m
c
m dv + (cv - mg) dt = 0
+
1
q
dt = 0
etc.
is not exact but
page 3 of solutions to review problems for Chapter 4
is exact (ºq/ºv = ºp/ºt = 0). The equation can be written as
d(
m
n(cv - mg) + t) = 0
c
Implicit solution is
m
n(cv - mg) + t = K
c
Solve for v to get explicit sol:
Kc
ct
m
m
Kc/m - ct/m
Kc/m -ct/m
cv - mg = e
= e
e
n(cv - mg) =
v =
-ct/m
= A e
mg
A -ct/m
+
e
c
c
3. At point (x,y) draw a little segment with slope x+y.
method 1 for solving
The equation is y' - y = x. This is linear first order with constant coeffs so
the methods of Chapter 1 work.
m = 1, yh = Aex.
Try yp = Bx + C. Substitute into the DE. You need B - (Bx + C) = x.
Equate the x coeffs: -B = c, B = -1.
Equate the constant terms: B - C = 0, C = -1.
So yp = -x-1, ygen = yh + yp = Aex - x - x.
method 2 for solving
The equation is linear. It can be written as y' - y = x.
-x dx = -xe-x - e-x + C.
P(x) = -1, Q(x) = x, e£P(x)dx = e-x, ye-x = ”xe
’
Solution is y = -x - 1 + Cex.
Yes you should be able to sketch the graph of this family. I would first sketch
y=-x-1 (a line) and y = Cex separately.
page 4 of solutions to review problems for Chapter 4
y=10e
x
1
y = 2 ex
1
y = - 2 ex
y=-x-1
y=-10e
x
Then add the line heights to each of the other curves. Way out to the left, Cex
is near 0 so the sum is like the line. Way out to the right, the exponential
heights are much larger in absolute value than the line heights so the sum is
like the exponential. When C = 0, the sum is the line.
Here are some of the curves in the family y = -x - 1 + ex, along with the
direction field.
C=2
C=0
C=-2
page 5 of solutions to review problems for Chapter 4
y
= A and then differentiate w.r.t. x on both sides.
x3
x-3 y' -3x-4 y = 0
3y
y' =
x
x
The orthogonal family has differential equation y' = . Now solve it.
3y
3y dy = - x dx
3y2
x2
= + C
2
2
x2 + 3y2 = 2C
x2 + 3y2 = K (an ellipse family)
4. Rewrite the DE as
A=1/2
9
K=
K=5
K=1
A=-4
5. A separable DE can be written as
(•)
x—stuff dx = y—stuff dy and rearranged to look like
(••)
x—stuff dx
p
Then
ºq
ºp
=
ºx
ºy
+
-y—stuff dy = 0
q
(both are 0) so (••) is exact.
To disprove the converse, i.e., to show that not every exact DE is also
separable, all you have to do is produce one counterexample. Problem 1(c) is
exact but not separable. QED