Internal
problem
ID
[4085] Book
:
Differential
equations,
Shepley
L.
Ross,
1964 Section
:
2.4,
page
55 Problem
number
:
10 Date
solved
:
Saturday, December 06, 2025 at 04:17:19 PM CAS
classification
:
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
2.1.10.1 Existence and uniqueness analysis
\begin{align*}
2 x +3 y \left (x \right )+1+\left (4 x +6 y \left (x \right )+1\right ) y^{\prime }\left (x \right )&=0 \\
y \left (-2\right ) &= 2 \\
\end{align*}
This is non linear first order ODE. In canonical form it is written as
\begin{align*} y^{\prime }\left (x \right ) &= f(x,y \left (x \right ))\\ &= -\frac {2 x +3 y +1}{4 x +6 y +1} \end{align*}
The \(x\) domain of \(f(x,y \left (x \right ))\) when \(y=2\) is
And
the point \(y_0 = 2\) is inside this domain. Now we will look at the continuity of
\begin{align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (-\frac {2 x +3 y +1}{4 x +6 y +1}\right ) \\ &= -\frac {3}{4 x +6 y +1}+\frac {12 x +18 y +6}{\left (4 x +6 y +1\right )^{2}} \end{align*}
The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=2\) is
This ODE is now solved for \(p \left (x \right )\).
No inversion is needed.
Integrating gives
\begin{align*} \int \frac {1}{\left (3 p +2\right ) \left (1+2 p \right )^{2}}d p &= dx\\ 3 \ln \left (3 p +2\right )-\frac {1}{1+2 p}-3 \ln \left (1+2 p \right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} \left (3 p +2\right ) \left (1+2 p \right )^{2}&= 0 \end{align*}
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular
solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (x \right ) = -{\frac {2}{3}}\\ p \left (x \right ) = -{\frac {1}{2}} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*}
2 x +3 y+1+\left (4 x +6 y+1\right ) y^{\prime }&=0 \\
y \left (-2\right ) &= 2 \\
\end{align*}
Applying
transformation
\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}
Results in the new ode which is Abel first kind
\begin{align*} 2 x +3 x u \left (x \right )+1+\left (4 x +6 x u \left (x \right )+1\right ) \left (u \left (x \right )+x u^{\prime }\left (x \right )\right ) = 0 \end{align*}
Which is now solved Unknown ode type.
2.1.10.3 Solved using first_order_ode_abel_second_kind_solved_by_converting_to_first_kind
0.430 (sec)This is Abel second kind ODE, it has the form
for \(u \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular
solution(s), which also have to satisfy the given ODE.
\begin{align*} u \left (x \right ) = 0\\ u \left (x \right ) = 2 \end{align*}
Substituting \(u \left (x \right )=\frac {1}{\frac {2 x}{3}+y \left (x \right )+\frac {1}{6}}\) in the above solution gives
Substituting equations (1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} \frac {A}{C x}-\frac {A x +B y}{C \,x^{2}}-\frac {\left (2 x +3 y +1\right ) B}{\left (4 x +6 y +1\right ) C x}+\frac {2}{4 x +6 y +1}-\frac {4 \left (2 x +3 y +1\right )}{\left (4 x +6 y +1\right )^{2}}-\frac {\left (-\frac {3}{4 x +6 y +1}+\frac {12 x +18 y +6}{\left (4 x +6 y +1\right )^{2}}\right ) \left (A x +B y \right )}{C x} = 0
\end{equation}
Putting the above in normal
form gives
\[
-\frac {8 B \,x^{3}+40 B \,x^{2} y +66 B x \,y^{2}+36 B \,y^{3}+3 A \,x^{2}+6 B \,x^{2}+20 B x y +12 B \,y^{2}+2 C \,x^{2}+B x +B y}{C \,x^{2} \left (4 x +6 y +1\right )^{2}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -8 B \,x^{3}-40 B \,x^{2} y -66 B x \,y^{2}-36 B \,y^{3}-3 A \,x^{2}-6 B \,x^{2}-20 B x y -12 B \,y^{2}-2 C \,x^{2}-B x -B y = 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
\[
\{x = v_{1}, y = v_{2}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} -8 B v_{1}^{3}-40 B v_{1}^{2} v_{2}-66 B v_{1} v_{2}^{2}-36 B v_{2}^{3}-3 A v_{1}^{2}-6 B v_{1}^{2}-20 B v_{1} v_{2}-12 B v_{2}^{2}-2 C v_{1}^{2}-B v_{1}-B v_{2} = 0
\end{equation}
Collecting the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}\}
\]
Equation (7E) now
becomes
\begin{equation}
\tag{8E} -8 B v_{1}^{3}-40 B v_{1}^{2} v_{2}+\left (-3 A -6 B -2 C \right ) v_{1}^{2}-66 B v_{1} v_{2}^{2}-20 B v_{1} v_{2}-B v_{1}-36 B v_{2}^{3}-12 B v_{2}^{2}-B v_{2} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
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 \(S(R)\). Therefore
\begin{align*} S &= \int { \frac {dx}{T}}\\ &= x \end{align*}
Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\)
are found, we need to setup the ode in these coordinates. This is done by evaluating
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= \frac {12 x +18 y +3}{2 x +3 y -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} &= \frac {18 R +3}{3 R -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 {\frac {18 R +3}{3 R -1}\, dR}\\ S \left (R \right ) &= 6 R +3 \ln \left (3 R -1\right ) + 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*} x = 4 x +6 y \left (x \right )+3 \ln \left (-1+2 x +3 y \left (x \right )\right )+c_2 \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 \((R,S)\)
\( \frac {dy}{dx} = -\frac {2 x +3 y +1}{4 x +6 y +1}\)
Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous C1storder, trying the canonical coordinates of the invariance group-> Calling odsolve with the ODE, diff(y(x),x) = -2/3, y(x)*** Sublevel 2 ***Methods for first order ODEs:--- Trying classification methods ---trying a quadraturetrying 1st order linear<- 1st order linear successful<-1st order, canonical coordinates successful<-homogeneous successful
2.1.10.6 ✓Mathematica. Time used: 2.368 (sec). Leaf size: 30