2.4.47 problem 44
Internal
problem
ID
[8362]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
44
Date
solved
:
Sunday, November 10, 2024 at 03:39:24 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} \cos \left (x \right ) y^{\prime \prime }+2 x y^{\prime }-x y&=0 \end{align*}
Using series expansion around \(x=0\)
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
\[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \]
Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change
of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let
initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives
\begin{align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\end{align*}
But
\begin{align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end{align}
And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as
\begin{align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6}\end{align}
Therefore (6) can be used from now on along with
\begin{equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7}\end{equation}
To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence
\begin{align*} F_0 &= -\frac {x \left (2 y^{\prime }-y\right )}{\cos \left (x \right )}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_{0}}{\partial x}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= -2 \left (\left (-2 x^{2} \sec \left (x \right )+x \tan \left (x \right )-\frac {x}{2}+1\right ) y^{\prime }+y \left (x^{2} \sec \left (x \right )-\frac {x \tan \left (x \right )}{2}-\frac {1}{2}\right )\right ) \sec \left (x \right )\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_{1}}{\partial x}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= 4 \sec \left (x \right )^{3} \left (\left (\frac {\left (1+x \right ) \cos \left (x \right )^{2}}{2}+\left (\left (-1+\frac {x}{2}\right ) \sin \left (x \right )-x^{2}+3 x \right ) \cos \left (x \right )-2 x^{3}+3 x^{2} \sin \left (x \right )-x \right ) y^{\prime }+\left (-\frac {x \cos \left (x \right )^{2}}{4}+\frac {\left (\frac {x^{2}}{2}-3 x +\sin \left (x \right )\right ) \cos \left (x \right )}{2}+x^{3}-\frac {3 x^{2} \sin \left (x \right )}{2}+\frac {x}{2}\right ) y\right )\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_{2}}{\partial x}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= -8 \sec \left (x \right )^{4} \left (\left (\left (\frac {3 x}{8}-\frac {3}{4}\right ) \cos \left (x \right )^{3}+\left (\left (-\frac {3}{4}-\frac {x}{4}\right ) \sin \left (x \right )+\frac {9 x}{4}-\frac {3}{2}+\frac {27 x^{2}}{8}\right ) \cos \left (x \right )^{2}+\left (\left (-7 x +\frac {9}{4} x^{2}\right ) \sin \left (x \right )+6 x^{2}-\frac {3 x}{4}-\frac {3 x^{3}}{2}+\frac {3}{2}\right ) \cos \left (x \right )-2 x \left (\left (-3 x^{2}-\frac {3}{4}\right ) \sin \left (x \right )+x^{3}+\frac {11 x}{4}\right )\right ) y^{\prime }+\left (\frac {3 \cos \left (x \right )^{3}}{8}+\left (-\frac {x}{2}-\frac {7 x^{2}}{4}+\frac {3}{4}+\frac {\sin \left (x \right ) x}{8}\right ) \cos \left (x \right )^{2}+\left (\left (\frac {7}{2} x -\frac {1}{2} x^{2}\right ) \sin \left (x \right )+\frac {x^{3}}{2}-3 x^{2}-\frac {3}{4}\right ) \cos \left (x \right )+x \left (\left (-3 x^{2}-\frac {3}{4}\right ) \sin \left (x \right )+x^{3}+\frac {11 x}{4}\right )\right ) y\right )\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_{3}}{\partial x}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= 16 \left (\left (\frac {\left (-3-\frac {x}{2}\right ) \cos \left (x \right )^{4}}{4}+\frac {\left (-3+\left (-\frac {x}{2}+1\right ) \sin \left (x \right )+7 x^{2}-\frac {87 x}{4}\right ) \cos \left (x \right )^{3}}{2}+\left (\left (-\frac {27}{8} x^{2}-7 x +5\right ) \sin \left (x \right )+\frac {3}{2}-5 x +\frac {97 x^{3}}{8}+6 x^{2}\right ) \cos \left (x \right )^{2}+\left (\left (6 x^{3}+\frac {3}{2} x -3-25 x^{2}\right ) \sin \left (x \right )-\frac {11 x^{2}}{2}+\frac {35 x}{2}-2 x^{4}+10 x^{3}\right ) \cos \left (x \right )+5 \left (\frac {5}{2} x^{2}+2 x^{4}\right ) \sin \left (x \right )-2 x^{5}-\frac {35 x^{3}}{2}-3 x \right ) y^{\prime }+\left (\frac {x \cos \left (x \right )^{4}}{16}+\frac {\left (1+\frac {45 x}{2}-\frac {11 x^{2}}{4}-\sin \left (x \right )\right ) \cos \left (x \right )^{3}}{4}+\frac {\left (\left (-5+\frac {15}{4} x^{2}+\frac {11}{4} x \right ) \sin \left (x \right )-\frac {99 x^{3}}{8}+5 x -\frac {15 x^{2}}{4}\right ) \cos \left (x \right )^{2}}{2}+\left (\frac {\left (3-\frac {15}{4} x^{3}+25 x^{2}\right ) \sin \left (x \right )}{2}-5 x^{3}+\frac {3 x^{4}}{4}-\frac {35 x}{4}+\frac {9 x^{2}}{8}\right ) \cos \left (x \right )+x \left (\frac {3}{2}+5 \left (-x^{3}-\frac {5}{4} x \right ) \sin \left (x \right )+x^{4}+\frac {35 x^{2}}{4}\right )\right ) y\right ) \sec \left (x \right )^{5} \end{align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives
\begin{align*} F_0 &= 0\\ F_1 &= -2 y^{\prime }\left (0\right )+y \left (0\right )\\ F_2 &= 2 y^{\prime }\left (0\right )\\ F_3 &= -3 y \left (0\right )+6 y^{\prime }\left (0\right )\\ F_4 &= -12 y^{\prime }\left (0\right )+4 y \left (0\right ) \end{align*}
Substituting all the above in (7) and simplifying gives the solution as
\[
y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) y \left (0\right )+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right )
\]
Since the expansion
point \(x = 0\) is an ordinary point, then this can also be solved using the standard power
series method. Let the solution be represented as power series of the form
\[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} \]
Then
\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} \end{align*}
Substituting the above back into the ode gives
\begin{align*} \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+2 x \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0\tag {1} \end{align*}
Expanding \(\cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives
\begin{align*} \cos \left (x \right ) &= 1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}-\frac {1}{720} x^{6} + \dots \\ &= 1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}-\frac {1}{720} x^{6} \end{align*}
Hence the ODE in Eq (1) becomes
\[
\left (1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}-\frac {1}{720} x^{6}\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+2 x \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0
\]
Expanding the first term in (1) gives
\[
\left (1\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )\right )-\frac {x^{2}}{2}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\frac {x^{4}}{24}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{6}}{720}\eslowast \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+2 x \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0
\]
Which simplifies to
\begin{equation}
\tag{2} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} a_{n}\right ) = 0
\end{equation}
The next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each
summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the power
and the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right ) \\
\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24} \\
\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) x^{n} \\
\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n}\right ) \\
\end{align*}
Substituting all the above in Eq (2) gives the following
equation where now all powers of \(x\) are the same and equal to \(n\).
\begin{equation}
\tag{3} \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n}\right ) = 0
\end{equation}
\(n=1\) gives
\[
6 a_{3}+2 a_{1}-a_{0}=0
\]
Which
after substituting earlier equations, simplifies to
\[
a_{3} = \frac {a_{0}}{6}-\frac {a_{1}}{3}
\]
\(n=2\) gives
\[
3 a_{2}+12 a_{4}-a_{1}=0
\]
Which after substituting
earlier equations, simplifies to
\[
a_{4} = \frac {a_{1}}{12}
\]
\(n=3\) gives
\[
3 a_{3}+20 a_{5}-a_{2} = 0
\]
Which after substituting earlier equations,
simplifies to
\[
\frac {a_{0}}{2}-a_{1}+20 a_{5} = 0
\]
Or
\[
a_{5} = -\frac {a_{0}}{40}+\frac {a_{1}}{20}
\]
\(n=4\) gives
\[
\frac {a_{2}}{12}+2 a_{4}+30 a_{6}-a_{3} = 0
\]
Which after substituting earlier equations, simplifies to
\[
\frac {a_{1}}{2}+30 a_{6}-\frac {a_{0}}{6} = 0
\]
Or
\[
a_{6} = \frac {a_{0}}{180}-\frac {a_{1}}{60}
\]
\(n=5\) gives
\[
\frac {a_{3}}{4}+42 a_{7}-a_{4} = 0
\]
Which after substituting earlier equations, simplifies to
\[
\frac {a_{0}}{24}-\frac {a_{1}}{6}+42 a_{7} = 0
\]
Or
\[
a_{7} = -\frac {a_{0}}{1008}+\frac {a_{1}}{252}
\]
For \(6\le n\), the
recurrence equation is
\begin{equation}
\tag{4} -\frac {\left (n -4\right ) a_{n -4} \left (n -5\right )}{720}+\frac {\left (n -2\right ) a_{n -2} \left (n -3\right )}{24}-\frac {n a_{n} \left (n -1\right )}{2}+\left (n +2\right ) a_{n +2} \left (1+n \right )+2 n a_{n}-a_{n -1} = 0
\end{equation}
Solving for \(a_{n +2}\), gives
\begin{align*}
\tag{5} a_{n +2}&= \frac {360 n^{2} a_{n}+n^{2} a_{n -4}-30 n^{2} a_{n -2}-1800 n a_{n}-9 n a_{n -4}+150 n a_{n -2}+20 a_{n -4}-180 a_{n -2}+720 a_{n -1}}{720 \left (n +2\right ) \left (1+n \right )} \\
&= \frac {\left (360 n^{2}-1800 n \right ) a_{n}}{720 \left (n +2\right ) \left (1+n \right )}+\frac {\left (n^{2}-9 n +20\right ) a_{n -4}}{720 \left (n +2\right ) \left (1+n \right )}+\frac {\left (-30 n^{2}+150 n -180\right ) a_{n -2}}{720 \left (n +2\right ) \left (1+n \right )}+\frac {a_{n -1}}{\left (n +2\right ) \left (1+n \right )} \\
\end{align*}
And so on. Therefore the solution is
\begin{align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ &= a_{3} x^{3}+a_{2} x^{2}+a_{1} x +a_{0} + \dots \end{align*}
Substituting the values for \(a_{n}\) found above, the solution becomes
\[
y = a_{0}+a_{1} x +\left (\frac {a_{0}}{6}-\frac {a_{1}}{3}\right ) x^{3}+\frac {a_{1} x^{4}}{12}+\left (-\frac {a_{0}}{40}+\frac {a_{1}}{20}\right ) x^{5}+\dots
\]
Collecting terms, the solution
becomes
\begin{equation}
\tag{3} y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) a_{0}+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) a_{1}+O\left (x^{6}\right )
\end{equation}
At \(x = 0\) the solution above becomes
\[
y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) c_1 +\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) c_2 +O\left (x^{6}\right )
\]
Maple step by step solution
Maple dsolve solution
Solving time : 0.008 (sec)
Leaf size : 44
dsolve(cos(x)*diff(diff(y(x),x),x)+2*diff(y(x),x)*x-x*y(x) = 0,y(x),
series,x=0)
\[
y = \left (1+\frac {1}{6} x^{3}-\frac {1}{40} x^{5}\right ) y \left (0\right )+\left (x -\frac {1}{3} x^{3}+\frac {1}{12} x^{4}+\frac {1}{20} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right )
\]
Mathematica DSolve solution
Solving time : 0.003 (sec)
Leaf size : 49
AsymptoticDSolveValue[{Cos[x]*D[y[x],{x,2}]+2*x*D[y[x],x]-x*y[x]==0,{}},
y[x],{x,0,5}]
\[
y(x)\to c_1 \left (-\frac {x^5}{40}+\frac {x^3}{6}+1\right )+c_2 \left (\frac {x^5}{20}+\frac {x^4}{12}-\frac {x^3}{3}+x\right )
\]