Solutions 2.6
2. Since ODE can be reduced to
y
2
y
0

1 = 0
,
this ODE is separable. We then and integrate this ODE in separable form:
Z
y
2
dt

Z
1
dt
=
C,
y
3
3

t
=
C.
The initial condition yields
y
(0)
3
3

1 =
C
or
8
3

1 =
C,
hence
C
= 5
/
3. Hence the solution to IVP in implicit form is
y
3
3

t
= 5
/
3
.
In explicit form, we have the solution
y
= (3
t
+ 5)
1
/
3
.
It exists for all
t
, but the slope becomes infinite at
t
=

5
/
3.
5. Since the ODE can be reduced to
y
0
y
3

t
= 0
,
this ODE is separable. Integration leads to

1
2
y
2

1
2
t
2
=
C,
and the initial condition yields

1
8
=
C.
1
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Hence the solution to IVP in implicit form is
1
2
y
2
+
1
2
t
2
=
1
8
.
In explicit form, we have
y
=
2
√
1

4
t
2
(choose the plus sign in front of the square root because of the initial condi
tion). This solution exists for

1
/
2
< t <
1
/
2.
9. Since this ODE can be reduced to
y
0
1

y
2

t
=
,
this ODE is separable. By integration
1
2
ln
1 +
y
1

y

t
2
2
=
C.
The initial condition yields
1
2
ln 3 =
C.
Thus the solution to IVP in implicit form is
ln
1 +
y
1

y

t
2
= ln 3
.
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 '06
 EDeSturler
 Calculus, Equations, Constant of integration, initial condition yields

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