Chapter 1
First-Order Differential
Equations
1.1 Terminology and Separable Equations
1. For x > 0, rewrite the equation as
2xy + 2y = ex .
With y = ϕ(x) = 12 x−1 (C − ex ), compute
1 −2
y = −x (C − ex ) − x−1 ex .
2
Then
2xy + 2y = x −x−2 (C − ex ) − x−1 ex + x−1 (C − ex ) = ex .
Therefore ϕ(x) is a solution.
2. For all x,
ϕ + ϕ = −Ce−x + (1 + Ce−x ) = 1
so ϕ(x) = 1 + Ce−x is a solution.
3. On any interval not containing x = 0 we have
2
1 3 3 x x −3
xϕ = x + 2 =x+ − =x− = x − ϕ,
2 2x 2x 2 2x
so ϕ is a solution.
4. With ϕ(x) = Ce−x ,
ϕ + ϕ = −Ce−x + Ce−x = 0,
so ϕ is a solution.
5. For x > 1,
√ 1
2ϕϕ = 2 x − 1 √ = 1,
2 x−1
so ϕ is a solution.
1
© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be
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, 2 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
√
6. For x = ± 2,
−2cx 2x c 2xϕ
ϕ = 2 = = ,
(x − 2)2 2 − x2 x2 − 2 2 − x2
so ϕ is a solution.
7. This equation is separable since we can write it as
sin(y) 1
dy = dx
cos(y) x
if cos(y) = 0 and x = 0. A routine integration gives the implicitly defined general solution
sec(y) = kx. Now cos(y) = 0 if y = (2n + 1)π/2 for n any integer. y = (2n + 1)π/2 also
satisfies the original differential equation and is a singular solution.
8. Substitute
sin(x − y) = sin(x) cos(y) − cos(x) sin(y),
cos(x + y) = cos(x) cos(y) − sin(x) sin(y),
and
cos(2x) = cos2 (x) − sin2 (x)
into the differential equation to obtain the separated equation
(cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx.
Upon integrating we obtain the implicitly defined solution
cos(y) + sin(y) = cos(x) + sin(x) + c.
9. This differential equation is not separable.
10. The differential equation itself assumes that y = 0 and x = −1. Write
x dy 2y 2 + 1
= ,
y dx x+1
which separates as
1 1
dy = dx.
y(2y 2 + 1) x(x + 1)
Use a partial fractions decomposition to write
1 2y 1 1
− dy = − dx.
y 1 + 2y 2 x 1+x
Integration this equation to obtain
1
ln |y| − ln(1 + 2y 2 ) = ln |x| − ln |x + 1| + c.
2
© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be
different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.
, 1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 3
Then,
y x
ln = ln + c,
1 + 2y 2 x+1
in which we have taken the case that y > 0 and x > 0 to drop the absolute values. Finally, take
the exponential of both sides of this equation to obtain the implicitly defined solution
y x
=k .
1 + 2y 2 x+1
Since y = 0 satisfies the original differential equation, y = 0 is a singular solution.
11. Write
dy 4x
3 = 2
dx y
and separate variables:
3y 2 dy = 4x dx.
Integrate to obtain
y 3 = 2x2 + k,
which implicitly defines the general solution. We can also write
1/3
y = 2x2 + k .
12. This equation is not separable.
13. Write the differential equation as
dy sin(x + y)
=
dx cos(y)
sin(x) cos(y) + cos(x) sin(y)
=
cos(y)
sin(y)
= sin(x) + cos(x) .
cos(y)
There is no way to separate the variables in this equation, so the differential equation is not
separable.
14. Since ex+y = ex ey , we can write the differential equation as
dy
ex ey = 3x
dx
or, in separated form,
ey dy = 3xe−x dx.
Integration gives us the implicitly defined general solution
ey = −3e−x (x + 1) + c.
© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be
different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.
, 4 CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS
15. Write the differential equation as
dy
x = y(y − 1).
dx
This is separable. If y = 0 and y = 1, we can write
1 1
dx = dy.
x y(y − 1)
Use partial fractions to write this as
1 1 1
dx = dy − dy.
x y−1 y
Integrate to obtain
ln |x| = ln |y − 1| − ln |y| + c,
or
y−1
ln |x| = ln + c.
y
This can be solved for x to obtain the general solution
1
y= .
1 − kx
The trivial solution y(x) = 0 is a singular solution, as is the constant solution y(x) = 1. We
assumed that y = 0, 1 in the algebra of separating the variables.
16. Write the differential equation as
dy
x = −y
dx
and separate the variables:
1 1
dy = − dx.
y x
This separation requires that x = 0 and y = 0. Integration gives us ln |y| = − ln |x| + c. Then
ln |y| + ln |x| = c
so ln |xy| = c. Then xy = ec = k, in which k can be any positive constant. Notice now that
y = 0 is also a solution of the original differential equation. Therefore, if we allow k to be
any constant (positive, negative or zero), we can omit the absolute values and write the general
solution in the implicit form xy = k.
17. Write ln(y x ) = x ln(y) and separate the variables to write
ln(y)
dy = 3x dx.
y
Integrate to obtain (ln(y))2 = 3x2 + c. Substitute the initial condition to obtain c = −3, so the
solution is implicitly defined by (ln(y))2 = 3x2 − 3.
© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be
different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.
