Internal
problem
ID
[1667] Book
:
Elementary
differential
equations
with
boundary
value
problems.
William
F.
Trench.
Brooks/Cole
2001 Section
:
Chapter
2,
First
order
equations.
Transformation
of
Nonlinear
Equations
into
Separable
Equations.
Section
2.4
Page
68 Problem
number
:
42 Date
solved
:
Thursday, October 10, 2024 at 05:00:52 PM CAS
classification
:
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
Solve
\begin{align*} y^{\prime }&=\frac {2 x +y+1}{x +2 y-4} \end{align*}
5.43.1 Solved as first order homogeneous class Maple C ode
Time used: 7.145 (sec)
Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)
An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous
functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if
\[ f(t^n X, t^n Y)= t^n f(X,Y) \]
In this
case, it can be seen that both \(M=2 X +Y\) and \(N=X +2 Y\) are both homogeneous and of the same order \(n=1\). Therefore
this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE
using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence
Which is now solved as separable in \(u \left (X \right )\).
The ode \(\frac {d}{d X}u \left (X \right ) = -\frac {2 \left (u \left (X \right )^{2}-1\right )}{X \left (2 u \left (X \right )+1\right )}\) is separable as it can be written as
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is
zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\frac {u^{2}-1}{2 u +1}=0\) for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=-1\\ u \left (X \right )&=1 \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
Therefore the solutions found are
\begin{align*} \frac {3 \ln \left (u \left (X \right )-1\right )}{2}+\frac {\ln \left (u \left (X \right )+1\right )}{2} = \ln \left (\frac {1}{X^{2}}\right )+c_1\\ u \left (X \right ) = -1\\ u \left (X \right ) = 1 \end{align*}
Converting \(\frac {3 \ln \left (u \left (X \right )-1\right )}{2}+\frac {\ln \left (u \left (X \right )+1\right )}{2} = \ln \left (\frac {1}{X^{2}}\right )+c_1\) back to \(Y \left (X \right )\) gives
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}+\frac {\left (2 x +y +1\right ) \left (b_{3}-a_{2}\right )}{x +2 y -4}-\frac {\left (2 x +y +1\right )^{2} a_{3}}{\left (x +2 y -4\right )^{2}}-\left (\frac {2}{x +2 y -4}-\frac {2 x +y +1}{\left (x +2 y -4\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {1}{x +2 y -4}-\frac {2 \left (2 x +y +1\right )}{\left (x +2 y -4\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {2 x^{2} a_{2}+4 x^{2} a_{3}-4 x^{2} b_{2}-2 x^{2} b_{3}+8 x y a_{2}+4 x y a_{3}-4 x y b_{2}-8 x y b_{3}+2 y^{2} a_{2}+4 y^{2} a_{3}-4 y^{2} b_{2}-2 y^{2} b_{3}-16 x a_{2}+4 x a_{3}-3 x b_{1}+2 x b_{2}+7 x b_{3}+3 y a_{1}-2 y a_{2}-7 y a_{3}+16 y b_{2}-4 y b_{3}-9 a_{1}-4 a_{2}+a_{3}-6 b_{1}-16 b_{2}+4 b_{3}}{\left (x +2 y -4\right )^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -2 x^{2} a_{2}-4 x^{2} a_{3}+4 x^{2} b_{2}+2 x^{2} b_{3}-8 x y a_{2}-4 x y a_{3}+4 x y b_{2}+8 x y b_{3}-2 y^{2} a_{2}-4 y^{2} a_{3}+4 y^{2} b_{2}+2 y^{2} b_{3}+16 x a_{2}-4 x a_{3}+3 x b_{1}-2 x b_{2}-7 x b_{3}-3 y a_{1}+2 y a_{2}+7 y a_{3}-16 y b_{2}+4 y b_{3}+9 a_{1}+4 a_{2}-a_{3}+6 b_{1}+16 b_{2}-4 b_{3} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= x +2 \\
\eta &= y -3 \\
\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 \\ &= y -3 - \left (\frac {2 x +y +1}{x +2 y -4}\right ) \left (x +2\right ) \\ &= \frac {-2 x^{2}+2 y^{2}-8 x -12 y +10}{x +2 y -4}\\ \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
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}{\frac {-2 x^{2}+2 y^{2}-8 x -12 y +10}{x +2 y -4}}} dy \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} &= 0 \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 \(f(R)\).
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = x +5\\ y = 1-x \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 ) = \frac {p \left (x \right )-\frac {-p \left (x \right )+2}{2 p \left (x \right )-1}}{-\frac {x}{2 p \left (x \right )-1}+\frac {2 x p \left (x \right )}{\left (2 p \left (x \right )-1\right )^{2}}-\frac {4 x}{\left (2 p \left (x \right )-1\right )^{2}}+\frac {4}{2 p \left (x \right )-1}-\frac {8 p \left (x \right )}{\left (2 p \left (x \right )-1\right )^{2}}-\frac {2}{\left (2 p \left (x \right )-1\right )^{2}}}\tag {3} \end{align*}
This ODE is now solved for \(p \left (x \right )\). No inversion is needed. The ode \(p^{\prime }\left (x \right ) = -\frac {2 \left (2 p \left (x \right )^{3}-p \left (x \right )^{2}-2 p \left (x \right )+1\right )}{3 \left (x +2\right )}\) is separable as it can be
written as
We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is
zero, since we had to divide by this above. Solving \(g(p)=0\) or \(-\frac {4}{3} p^{3}+\frac {2}{3} p^{2}+\frac {4}{3} p -\frac {2}{3}=0\) for \(p \left (x \right )\) gives
\begin{align*} p \left (x \right )&=-1\\ p \left (x \right )&=1\\ p \left (x \right )&={\frac {1}{2}} \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.