PRACTICE PROBLEMS: SET 8
MATH 101: PROF. DRAGOS GHIOCA
1.
Problems
Problem 1.
Solve the following ordinary differential equations:
(a)
(b) y
′
= y 2 − y
; general solution for x > 0.
x y
′
=
√ xy ; general solution for x > 0.
(c) y
′
= xe y with initial condition y (1) =
−
5 .
(d) y
′
= x y 2 with initial condition y (0) = 2 .
1

2 MATH 101: PROF. DRAGOS GHIOCA
2.
Solutions
Problem 1.
Each time we isolate on different sides of the equation the functions in y , respectively the functions in x . Each time, in order to integrate, it helps to represent y
′
( x ) as dy dx
.
(a) We obtain y 2 dy
− y
= dx
, x and thus ∫ y 2 dy
− y
=
∫ dx x
.
We use partial fractions for the integral on the left-hand side and obtain y 2
1
− y
=
1 y
−
1
−
1 y
, by easy algebra. Therefore
∫
1 y
−
1
−
1 y dy =
∫
1 x dx , which yields ln
| y
−
1
| − ln
| y
|
= ln
| x
|
+
C
.
Hence ln y
−
1 y
= ln
| x
|
+ and thus after exponentiating we get y
−
1 y
=
K ·| x
|
,
C
, where
K
= e
C is a positive constant. Since we are interested only in solution for x > 0, we are allowed to drop the absolute value on the right-hand side.
We drop the absolute value on the left-hand side at the price of allowing
K be arbitrary nonzero real number (not necessarily positive). Hence y − 1 y
=
K x, and we solve for y in terms of x . We obtain: y ( x ) =
1
1
− K x
, for an arbitrary nonzero real number
K
.
Note that K = 0 is indeed excluded since otherwise y ( x ) = 1 and then the differential equation wouldn’t make sense; y ( x ) cannot be either 0 or 1 due to the presence of y 2 − y in the denominator. The fact that y ( x ) = 0 comes directly from the formula we obtained for y ( x ). Finally, note that the above formula for y ( x ) does not make sense for defined as a differentiable function either on (0 , to avoid the point where y ( x ) is not well-defined.
1
K x =
) or on (
1
K
1
K
, i.e.
y ( x ) is
, ∞ ) in order

PRACTICE PROBLEMS: SET 8 3
(b) We have and thus ∫
′
=
√ x, y y
− 1
2 dy =
∫ x
1
2 dx .
After integrating we get
2 y
1
2 =
2
3 x
3
2 +
C
, and thus y
1
2
= x
3
2
+ K ,
3 after dividing by 2 and absorbing the factor square both sides and conclude that x 3 y ( x ) =
9 for some arbitrary constant
K
.
+
(c) We have y
′
3
= x, e y
2
K x
3
2 and so, ∫ ∫ e
− y dy = x
+ dx .
K 2
,
1
2 into the constant
K
. We
Therefore x
2
− e
− y
= +
C
,
2 and thus solving for y , we obtain y ( x ) =
− ln
(
K − x 2
2
)
, where
K
=
− C is simply a constant. Now we use the information given by y (1) =
−
5 and thus conclude that
−
5 =
− ln
(
K −
1
2
)
.
We solve for the constant
K and obtain
K
= e
5
+
1
2
; hence y ( x ) = − ln
( e
5
+
1 − x 2
)
.
2
We note that y ( x ) is well-defined whenever the quantity inside the natural logarithm is positive. Therefore the domain of definition for y ( x ) is limited by i.e., x 2 −
1
< e
5
,
2
−
√
2 e 5 + 1 < x <
√
2 e 5 + 1 .

4 MATH 101: PROF. DRAGOS GHIOCA
(d) We have and so, ∫ y
′ y
2
= x, y
2 dy =
∫ x dx .
Therefore y 3 x 2
= +
C
,
3 2 and so, √
3 x 2 y ( x ) =
3
+
K
,
2 where the factor 3 was absorbed into the constant K . Then we use the initial condition y (0) = 2 and conclude that
K
= 8 .
In conclusion, y ( x ) =
√
3
3 x 2
+ 8 .
2