Let
P
(
x
,
y
)
=
(
x
+
y
)
k
{\displaystyle P(x,y)=(x+y)^{k}}
polynomial of two variables of order
k
{\displaystyle k}
k
{\displaystyle k}
natural number . By the binomial formula ,
P
(
x
,
y
)
=
∑
j
=
0
k
(
k
j
)
x
j
y
k
−
j
{\displaystyle P(x,y)=\sum \limits _{j=0}^{k}{{\binom {k}{j}}x^{j}y^{k-j}}}
[relevant? The binomial differential equation becomes
(
y
′
)
m
=
(
x
+
y
)
k
{\textstyle (y')^{m}=(x+y)^{k}}
[clarification needed Substituting
v
=
x
+
y
{\displaystyle v=x+y}
v
′
=
1
+
y
′
{\displaystyle v'=1+y'}
(
v
′
−
1
)
m
=
v
k
{\textstyle (v'-1)^{m}=v^{k}}
d
v
d
x
=
1
+
v
k
m
{\textstyle {\tfrac {dv}{dx}}=1+v^{\tfrac {k}{m}}}
separable ordinary differential equation. Solving gives
d
v
d
x
=
1
+
v
k
m
⇒
d
v
1
+
v
k
m
=
d
x
⇒
∫
d
v
1
+
v
k
m
=
x
+
C
{\displaystyle {\begin{array}{lrl}&{\frac {dv}{dx}}&=1+v^{\tfrac {k}{m}}\\\Rightarrow &{\frac {dv}{1+v^{\tfrac {k}{m}}}}&=dx\\\Rightarrow &\int {\frac {dv}{1+v^{\tfrac {k}{m}}}}&=x+C\end{array}}}
If
m
=
k
{\displaystyle m=k}
v
′
−
1
=
v
{\displaystyle v'-1=v}
y
(
x
)
=
C
e
x
−
x
−
1
{\displaystyle y\left(x\right)=Ce^{x}-x-1}
C
{\displaystyle C}
If
m
|
k
{\displaystyle m|k}
m
{\displaystyle m}
divisor of
k
{\displaystyle k}
∫
d
v
1
+
v
n
=
x
+
C
{\textstyle \int {\frac {dv}{1+v^{n}}}=x+C}
Gradshteyn and Ryzhik
∫
d
v
1
+
v
n
=
{
−
2
n
∑
i
=
0
n
2
−
1
P
i
cos
(
2
i
+
1
n
π
)
+
2
n
∑
i
=
0
n
2
−
1
Q
i
sin
(
2
i
+
1
n
π
)
,
n
:
even integer
1
n
ln
(
1
+
v
)
−
2
n
∑
i
=
0
n
−
3
2
P
i
cos
(
2
i
+
1
n
π
)
+
2
n
∑
i
=
0
n
−
3
2
Q
i
sin
(
2
i
+
1
n
π
)
,
n
:
odd integer
{\displaystyle \int {\frac {dv}{1+v^{n}}}=\left\{{\begin{array}{ll}-{\frac {2}{n}}\sum \limits _{i=0}^{{\textstyle {n \over 2}}-1}{P_{i}\cos \left({{\frac {2i+1}{n}}\pi }\right)}+{\frac {2}{n}}\sum \limits _{i=0}^{{\tfrac {n}{2}}-1}{Q_{i}\sin \left({{\frac {2i+1}{n}}\pi }\right)},&n:{\text{even integer}}\\\\{\frac {1}{n}}\ln \left({1+v}\right)-{\frac {2}{n}}\sum \limits _{i=0}^{\textstyle {{n-3} \over 2}}{P_{i}\cos \left({{\frac {2i+1}{n}}\pi }\right)}+{\frac {2}{n}}\sum \limits _{i=0}^{\tfrac {n-3}{2}}{Q_{i}\sin \left({{\frac {2i+1}{n}}\pi }\right)},&n:{\text{odd integer}}\\\end{array}}\right.}
where
P
i
=
1
2
ln
(
v
2
−
2
v
cos
(
2
i
+
1
n
π
)
+
1
)
Q
i
=
arctan
(
v
−
cos
(
2
i
+
1
n
π
)
sin
(
2
i
+
1
n
π
)
)
{\displaystyle {\begin{aligned}P_{i}&={\frac {1}{2}}\ln \left({v^{2}-2v\cos \left({{\frac {2i+1}{n}}\pi }\right)+1}\right)\\Q_{i}&=\arctan \left({\frac {v-\cos \left({{\textstyle {{2i+1} \over n}}\pi }\right)}{\sin \left({{\textstyle {{2i+1} \over n}}\pi }\right)}}\right)\end{aligned}}}
