2.1.59 problem 59
Internal
problem
ID
[8197]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
59
Date
solved
:
Sunday, November 10, 2024 at 03:09:37 AM
CAS
classification
:
[[_homogeneous, `class C`], _dAlembert]
Solve
\begin{align*} y^{\prime }+\sin \left (-y+x \right )&=0 \end{align*}
Time used: 0.193 (sec)
The given ode has the general form
\begin{align*} y^{\prime } & = B+C f\left ( ax +b y +c\right ) \tag {1} \end{align*}
Comparing (1) to the ode given shows the parameters in the ODE have these values
\begin{align*} B &= 0\\ C &= -1\\ a &= 1\\ b &= -1\\ c &= 0 \end{align*}
This form of ode can be solved by change of variables \(u=ax+b y +c\) which makes the ode separable.
\begin{align*} u^{\prime }\left (x \right ) &=a+b y^{\prime } \end{align*}
Or
\begin{align*} y^{\prime } &= \frac { u^{\prime }\left (x \right ) - a} {b} \end{align*}
The ode becomes
\begin{align*} \frac {u' - a}{b} & = B+C f\left ( u\right ) \\ u' & =b B+ b C f\left ( u\right ) +a \\ \frac {du}{b B+b C f\left ( u\right ) +a} &= d x \end{align*}
Integrating gives
\begin{align*} \int \frac {du}{b B+ b C f(u) +a} &=x+c_1\\ \int ^{u}\frac {d\tau }{b B + b C f(\tau ) +a} & = x+c_1 \end{align*}
Replacing back \(u=ax+b y +c\) the above becomes
\begin{equation} \int ^{ax+b y +c}\frac {d\tau }{b B+b C f\left ( \tau \right ) +a} = x+c_{1}\tag {2} \end{equation}
If initial conditions are given as \(y\left ( x_{0}\right ) = y_{0}\) , the above becomes
\begin{align*} \int _{0}^{a x_{0}+b y_{0}+c}\frac {d\tau }{b B + b C f\left ( \tau \right ) +a} & =x_{0}+c_{1}\\ c_{1} & =\int _{0}^{ax+by_{0}+c}\frac {d\tau }{b B+ b C f\left ( \tau \right )+a}-x_{0} \end{align*}
Substituting this into (2) gives
\begin{align*} \int ^{ax+by+c}\frac {d\tau }{bB+bC f\left ( \tau \right ) +a}= x+\int _{0}^{ax+by_{0}+c}\frac {d\tau }{bB+bC f\left ( \tau \right ) +a}-x_{0} \tag {3} \end{align*}
Since no initial conditions are given, then using (2) and replacing the values of the
parameters into (2) gives the solution as
\[
\int _{}^{-y+x}\frac {1}{1+\sin \left (\tau \right )}d \tau = x +c_1
\]
Which simplifies to
\[
-\frac {2}{\tan \left (-\frac {y}{2}+\frac {x}{2}\right )+1} = x +c_1
\]
Solving for \(y\) gives
\begin{align*}
y &= x +2 \arctan \left (\frac {c_1 +x +2}{x +c_1}\right ) \\
\end{align*}
Figure 2.142: Slope field plot
\(y^{\prime }+\sin \left (-y+x \right ) = 0\)
Summary of solutions found
\begin{align*}
y &= x +2 \arctan \left (\frac {c_1 +x +2}{x +c_1}\right ) \\
\end{align*}
Solved using Lie symmetry for first order ode
Time used: 0.726 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=-\sin \left (-y +x \right )\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to
use as anstaz gives
\begin{align*}
\tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\
\tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\
\end{align*}
Where the unknown coefficients are
\[
\{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\}
\]
Substituting equations (1E,2E) and \(\omega \)
into (A) gives
\begin{equation}
\tag{5E} b_{2}-\sin \left (-y +x \right ) \left (b_{3}-a_{2}\right )-\sin \left (-y +x \right )^{2} a_{3}+\cos \left (-y +x \right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\cos \left (-y +x \right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\sin \left (-y +x \right )^{2} a_{3}+\cos \left (-y +x \right ) x a_{2}-\cos \left (-y +x \right ) x b_{2}+\cos \left (-y +x \right ) y a_{3}-\cos \left (-y +x \right ) y b_{3}+\sin \left (-y +x \right ) a_{2}-\sin \left (-y +x \right ) b_{3}+\cos \left (-y +x \right ) a_{1}-\cos \left (-y +x \right ) b_{1}+b_{2} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -\sin \left (-y +x \right )^{2} a_{3}+\cos \left (-y +x \right ) x a_{2}-\cos \left (-y +x \right ) x b_{2}+\cos \left (-y +x \right ) y a_{3}-\cos \left (-y +x \right ) y b_{3}+\sin \left (-y +x \right ) a_{2}-\sin \left (-y +x \right ) b_{3}+\cos \left (-y +x \right ) a_{1}-\cos \left (-y +x \right ) b_{1}+b_{2} = 0
\end{equation}
Simplifying the above gives
\begin{equation}
\tag{6E} b_{2}-\frac {a_{3}}{2}+\frac {a_{3} \cos \left (-2 y +2 x \right )}{2}+\cos \left (-y +x \right ) x a_{2}-\cos \left (-y +x \right ) x b_{2}+\cos \left (-y +x \right ) y a_{3}-\cos \left (-y +x \right ) y b_{3}+\sin \left (-y +x \right ) a_{2}-\sin \left (-y +x \right ) b_{3}+\cos \left (-y +x \right ) a_{1}-\cos \left (-y +x \right ) b_{1} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y, \cos \left (-2 y +2 x \right ), \cos \left (-y +x \right ), \sin \left (-y +x \right )\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\{x = v_{1}, y = v_{2}, \cos \left (-2 y +2 x \right ) = v_{3}, \cos \left (-y +x \right ) = v_{4}, \sin \left (-y +x \right ) = v_{5}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} b_{2}-\frac {1}{2} a_{3}+\frac {1}{2} a_{3} v_{3}+v_{4} v_{1} a_{2}-v_{4} v_{1} b_{2}+v_{4} v_{2} a_{3}-v_{4} v_{2} b_{3}+v_{5} a_{2}-v_{5} b_{3}+v_{4} a_{1}-v_{4} b_{1} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} b_{2}-\frac {a_{3}}{2}+\left (a_{2}-b_{2}\right ) v_{1} v_{4}+\left (a_{3}-b_{3}\right ) v_{2} v_{4}+\frac {a_{3} v_{3}}{2}+\left (a_{1}-b_{1}\right ) v_{4}+\left (a_{2}-b_{3}\right ) v_{5} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} \frac {a_{3}}{2}&=0\\ a_{1}-b_{1}&=0\\ a_{2}-b_{2}&=0\\ a_{2}-b_{3}&=0\\ a_{3}-b_{3}&=0\\ b_{2}-\frac {a_{3}}{2}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=b_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=0 \end{align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= 1 \\
\eta &= 1 \\
\end{align*}
Shifting is now applied to make \(\xi =0\) in order to simplify the rest of
the computation
\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= 1 - \left (-\sin \left (-y +x \right )\right ) \left (1\right ) \\ &= 1-\sin \left (y \right ) \cos \left (x \right )+\cos \left (y \right ) \sin \left (x \right )\\ \xi &= 0 \end{align*}
The next step is to determine the canonical coordinates \(R,S\) . The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\)
where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and
hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\) . Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\) . Since
\(\xi =0\) then in this special case
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{1-\sin \left (y \right ) \cos \left (x \right )+\cos \left (y \right ) \sin \left (x \right )}} dy \end{align*}
Which results in
\begin{align*} S&= -\frac {2}{-\tan \left (-\frac {y}{2}+\frac {x}{2}\right )-1} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode
given by
\begin{align*} \omega (x,y) &= -\sin \left (-y +x \right ) \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {1}{1+\sin \left (-y +x \right )}\\ S_{y} &= \frac {1}{1+\sin \left (-y +x \right )} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -1\tag {2A} \end{align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of
\(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= -1 \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts
an ode, no matter how complicated it is, to one that can be solved by integration when the
ode is in the canonical coordiates \(R,S\) .
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\) , then we only need to integrate \(f(R)\) .
\begin{align*} \int {dS} &= \int {-1\, dR}\\ S \left (R \right ) &= -R + c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} \frac {2}{\tan \left (-\frac {y}{2}+\frac {x}{2}\right )+1} = -x +c_2 \end{align*}
Which gives
\begin{align*} y = x +2 \arctan \left (\frac {c_2 -x -2}{-x +c_2}\right ) \end{align*}
The following diagram shows solution curves of the original ode and how they transform in
the canonical coordinates space using the mapping shown.
Original ode in \(x,y\) coordinates
Canonical
coordinates
transformation
ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = -\sin \left (-y +x \right )\)
\( \frac {d S}{d R} = -1\)
\(\!\begin {aligned} R&= x\\ S&= \frac {2}{\tan \left (-\frac {y}{2}+\frac {x}{2}\right )+1} \end {aligned} \)
Figure 2.143: Slope field plot
\(y^{\prime }+\sin \left (-y+x \right ) = 0\)
Summary of solutions found
\begin{align*}
y &= x +2 \arctan \left (\frac {c_2 -x -2}{-x +c_2}\right ) \\
\end{align*}
Solved as first order ode of type dAlembert
Time used: 0.364 (sec)
Let \(p=y^{\prime }\) the ode becomes
\begin{align*} p +\sin \left (-y +x \right ) = 0 \end{align*}
Solving for \(y\) from the above results in
\begin{align*}
\tag{1} y &= x +\arcsin \left (p \right ) \\
\end{align*}
This has the form
\begin{align*} y=xf(p)+g(p)\tag {*} \end{align*}
Where \(f,g\) are functions of \(p=y'(x)\) . The above ode is dAlembert ode which is now solved.
Taking derivative of (*) w.r.t. \(x\) gives
\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}
Comparing the form \(y=x f + g\) to (1A) shows that
\begin{align*} f &= 1\\ g &= \arcsin \left (p \right ) \end{align*}
Hence (2) becomes
\begin{align*} p -1 = \frac {p^{\prime }\left (x \right )}{\sqrt {-p^{2}+1}}\tag {2A} \end{align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
\begin{align*} p -1 = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=1 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = x +\frac {\pi }{2} \end{align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\) . From eq. (2A). This results in
\begin{align*} p^{\prime }\left (x \right ) = \left (p \left (x \right )-1\right ) \sqrt {-p \left (x \right )^{2}+1}\tag {3} \end{align*}
Inverting the above ode gives
\begin{align*} \frac {d}{d p}x \left (p \right ) = \frac {1}{\sqrt {-p^{2}+1}\, \left (p -1\right )}\tag {4} \end{align*}
This ODE is now solved for \(x \left (p \right )\) . Integrating gives
\begin{align*} x \left (p \right )&=\int \frac {1}{\sqrt {-p^{2}+1}\, \left (p -1\right )}d p +c_1\\ x \left (p \right )&=-\frac {p +1}{\sqrt {-p^{2}+1}}+c_1\tag {5} \end{align*}
Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve
for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\)
from Eq. (5) and substitute the result into (1A).
Eliminating \(p\) from the following two equations
\begin{align*}
x &= -\frac {p +1}{\sqrt {-p^{2}+1}}+c_1 \\
y &= x +\arcsin \left (p \right ) \\
\end{align*}
results in
\begin{align*}
p &= \frac {c_1^{2}-2 c_1 x +x^{2}-1}{c_1^{2}-2 c_1 x +x^{2}+1} \\
\end{align*}
Substituting the above into Eq
(1A) and simplifying gives
\begin{align*}
y &= x +\arcsin \left (\frac {c_1^{2}-2 c_1 x +x^{2}-1}{c_1^{2}-2 c_1 x +x^{2}+1}\right ) \\
\end{align*}
Figure 2.144: Slope field plot
\(y^{\prime }+\sin \left (-y+x \right ) = 0\)
Summary of solutions found
\begin{align*}
y &= x +\frac {\pi }{2} \\
y &= x +\arcsin \left (\frac {c_1^{2}-2 c_1 x +x^{2}-1}{c_1^{2}-2 c_1 x +x^{2}+1}\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+\sin \left (x -y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sin \left (x -y \left (x \right )\right ) \end {array} \]
Maple trace
` Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
trying inverse linear
trying homogeneous types:
trying homogeneous C
1 st order, trying the canonical coordinates of the invariance group
<- 1st order, canonical coordinates successful
<- homogeneous successful `
Maple dsolve solution
Solving time : 0.046
(sec)
Leaf size : 23
dsolve ( diff ( y ( x ), x )+ sin ( x - y ( x )) = 0,
y(x),singsol=all)
\[
y = x +2 \arctan \left (\frac {c_{1} -x -2}{-x +c_{1}}\right )
\]
Mathematica DSolve solution
Solving time : 37.899
(sec)
Leaf size : 553
DSolve [{ D [ y [ x ], x ]- Sin [ y [ x ]- x ]==0,{}},
y[x],x,IncludeSingularSolutions-> True ]
\begin{align*}
y(x)\to -2 \arccos \left (\frac {(-x+2+c_1) \cos \left (\frac {x}{2}\right )+(x-c_1) \sin \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 (1+c_1) x+2+c_1{}^2+2 c_1}}\right ) \\
y(x)\to 2 \arccos \left (\frac {(-x+2+c_1) \cos \left (\frac {x}{2}\right )+(x-c_1) \sin \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 (1+c_1) x+2+c_1{}^2+2 c_1}}\right ) \\
y(x)\to -2 \arccos \left (\frac {(x-2-c_1) \cos \left (\frac {x}{2}\right )+(-x+c_1) \sin \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 (1+c_1) x+2+c_1{}^2+2 c_1}}\right ) \\
y(x)\to 2 \arccos \left (\frac {(x-2-c_1) \cos \left (\frac {x}{2}\right )+(-x+c_1) \sin \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 (1+c_1) x+2+c_1{}^2+2 c_1}}\right ) \\
y(x)\to -2 \arccos \left (\frac {\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \\
y(x)\to 2 \arccos \left (\frac {\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \\
y(x)\to -2 \arccos \left (\frac {\sin \left (\frac {x}{2}\right )-\cos \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \\
y(x)\to 2 \arccos \left (\frac {\sin \left (\frac {x}{2}\right )-\cos \left (\frac {x}{2}\right )}{\sqrt {2}}\right ) \\
y(x)\to -2 \arccos \left (\frac {(x-2) \cos \left (\frac {x}{2}\right )-x \sin \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 x+2}}\right ) \\
y(x)\to 2 \arccos \left (\frac {(x-2) \cos \left (\frac {x}{2}\right )-x \sin \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 x+2}}\right ) \\
y(x)\to -2 \arccos \left (\frac {x \sin \left (\frac {x}{2}\right )-(x-2) \cos \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 x+2}}\right ) \\
y(x)\to 2 \arccos \left (\frac {x \sin \left (\frac {x}{2}\right )-(x-2) \cos \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {x^2-2 x+2}}\right ) \\
\end{align*}