[<< wikibooks] Ordinary Differential Equations/Separable 4
== Existence problems ==
1) f(x,y) has no discontinuities, so a solution exists.

∂

f

∂

y

{\displaystyle {\frac {\partial {f}}{\partial {y}}}}
has no discontinuities, so the solution is unique.
2) f(x,y) is not defined for the point (-1,10) because ln(x) is not defined.  So no solution exists.
3) f(x,y) has discontinuities at y=1 and -1, but not at 0 so a solution exists.

∂

f

∂

y

{\displaystyle {\frac {\partial {f}}{\partial {y}}}}
has no discontinuities at (0,16) so the solution is unique.
4) f(x,y) has discontinuities at y<0, but not at 1 so a solution exists.

∂

f

∂

y

{\displaystyle {\frac {\partial {f}}{\partial {y}}}}
is discontinuous at 1, so the solution is not unique
5) f(x,y) has discontinuities at -3 and -4, but not at 0 so a solution exists.

∂

f

∂

y

{\displaystyle {\frac {\partial {f}}{\partial {y}}}}
has no discontinuities at (5,9) so the solution is unique.
6) f(x,y) has a discontinuity at x=5, so no solution exists.

== Separable equations ==
7)

y
′

=

y

3

s
e

c

2

(
x
)

{\displaystyle y'=y^{3}sec^{2}(x)}

d
y

y

3

=
s
e

c

2

(
x
)
d
x

{\displaystyle {\frac {dy}{y^{3}}}=sec^{2}(x)dx}

∫

d
y

y

3

=
∫
s
e

c

2

(
x
)
d
x

{\displaystyle \int {\frac {dy}{y^{3}}}=\int sec^{2}(x)dx}

−

1

2

y

2

=
t
a
n
(
x
)
+
C

{\displaystyle -{\frac {1}{2y^{2}}}=tan(x)+C}

y
=
−

1

(
2
t
a
n
(
x
)
+
C
)

{\displaystyle y=-{\frac {1}{\sqrt {(2tan(x)+C)}}}}

8)

y
′

=

5

y

2

+
6

y

{\displaystyle y'={\frac {5y^{2}+6}{y}}}

y
d
y

5

y

2

+
6

=
d
x

{\displaystyle {\frac {ydy}{5y^{2}+6}}=dx}

∫

y
d
y

5

y

2

+
6

=
∫
d
x

{\displaystyle \int {\frac {ydy}{5y^{2}+6}}=\int dx}

1
10

l
n
(
5

y

2

+
6
)
=
x
+
C

{\displaystyle {\frac {1}{10}}ln(5y^{2}+6)=x+C}

y
=
±

C

e

10
x

−

6
5

{\displaystyle y=\pm {\sqrt {Ce^{10x}-{\frac {6}{5}}}}}

9)

y
′

=

x

3

/

y

3

{\displaystyle y'=x^{3}/y^{3}}

y

3

d
y
=

x

3

d
x

{\displaystyle y^{3}dy=x^{3}dx}

∫

y

3

d
y
=
∫

x

3

d
x

{\displaystyle \int y^{3}dy=\int x^{3}dx}

1
4

y

4

=

1
4

x

4

+
C

{\displaystyle {\frac {1}{4}}y^{4}={\frac {1}{4}}x^{4}+C}

y
=
(

x

4

+
C

)

1
4

{\displaystyle y=(x^{4}+C)^{\frac {1}{4}}}

10)

y
′

=

x

2

+
3
x
−
9

{\displaystyle y'=x^{2}+3x-9}

d
y
=
(

x

2

+
3
x
−
9
)
d
x

{\displaystyle dy=(x^{2}+3x-9)dx}

∫
d
y
=
∫
(

x

2

+
3
x
−
9
)
d
x

{\displaystyle \int dy=\int (x^{2}+3x-9)dx}

y
=

1
3

x

3

+

3
2

x

2

−
9
x
+
C

{\displaystyle y={\frac {1}{3}}x^{3}+{\frac {3}{2}}x^{2}-9x+C}

11)

y
′

=
c
o
s
(
y
)

/

s
i
n
(
y
)

{\displaystyle y'=cos(y)/sin(y)}

s
i
n
(
y
)
d
y

c
o
s
(
y
)

=
d
x

{\displaystyle {\frac {sin(y)dy}{cos(y)}}=dx}

∫

s
i
n
(
y
)
d
y

c
o
s
(
y
)

=
∫
d
x

{\displaystyle \int {\frac {sin(y)dy}{cos(y)}}=\int dx}

−
l
n
(
c
o
s
(
y
)
)
=
x
+
C

{\displaystyle -ln(cos(y))=x+C}

y
=
a
r
c
c
o
s
(
C

e

x

)

{\displaystyle y=arccos(Ce^{x})}

12)

y
′

=

c
o
s
(
x
)

s
i
n
(
y
)

{\displaystyle y'={\frac {cos(x)}{sin(y)}}}

s
i
n
(
y
)
d
y
=
c
o
s
(
x
)
d
x

{\displaystyle sin(y)dy=cos(x)dx}

∫
s
i
n
(
y
)
d
y
=
∫
c
o
s
(
x
)
d
x

{\displaystyle \int sin(y)dy=\int cos(x)dx}

−
c
o
s
(
y
)
=
s
i
n
(
x
)
+
C

{\displaystyle -cos(y)=sin(x)+C}

y
=
a
r
c
c
o
s
(
−
s
i
n
(
x
)
+
C
)

{\displaystyle y=arccos(-sin(x)+C)}

== Initial value problems ==
13)

y
′

=
c
o
s
(
x
)
+
s
i
n
(
x
)
,
y
(
0
)
=
1

{\displaystyle y'=cos(x)+sin(x),y(0)=1}

d
y
=
(
c
o
s
(
x
)
+
s
i
n
(
x
)
)
d
x

{\displaystyle dy=(cos(x)+sin(x))dx}

∫
d
y
=
∫
(
c
o
s
(
x
)
+
s
i
n
(
x
)
)
d
x

{\displaystyle \int dy=\int (cos(x)+sin(x))dx}

y
=
s
i
n
(
x
)
−
c
o
s
(
x
)
+
C

{\displaystyle y=sin(x)-cos(x)+C}

1
=
s
i
n
(
0
)
−
c
o
s
(
0
)
+
C
=
0
−
1
+
C
=
C
−
1

{\displaystyle 1=sin(0)-cos(0)+C=0-1+C=C-1}

C
=
2

{\displaystyle C=2}

y
=
s
i
n
(
x
)
−
c
o
s
(
x
)
+
2

{\displaystyle y=sin(x)-cos(x)+2}

14)

y
′

=
7

y

2

,
y
(
5
)
=
9

{\displaystyle y'=7y^{2},y(5)=9}

d
y

y

2

=
7
d
x

{\displaystyle {\frac {dy}{y^{2}}}=7dx}

∫

d
y

y

2

=
∫
7
d
x

{\displaystyle \int {\frac {dy}{y^{2}}}=\int 7dx}

−

1
y

=
7
x
+
C

{\displaystyle -{\frac {1}{y}}=7x+C}

y
=

1

−
7
x
+
C

{\displaystyle y={\frac {1}{-7x+C}}}

9
=

1

−
7
∗
5
+
C

{\displaystyle 9={\frac {1}{-7*5+C}}}

C
=

316
9

{\displaystyle C={\frac {316}{9}}}

y
=

1

−
7
x
+

316
9

{\displaystyle y={\frac {1}{-7x+{\frac {316}{9}}}}}