Internal
problem
ID
[4081] Book
:
Differential
equations,
Shepley
L.
Ross,
1964 Section
:
2.4,
page
55 Problem
number
:
6 Date
solved
:
Wednesday, November 26, 2025 at 03:32:38 PM CAS
classification
:
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
Solved using first_order_ode_dAlembert
Time used: 0.218 (sec)
Solve
\begin{align*}
3 x -y+1-\left (6 x -2 y-3\right ) y^{\prime }&=0 \\
\end{align*}
Let \(p=y^{\prime }\) the ode becomes
\begin{align*} 3 x -y +1-\left (6 x -2 y -3\right ) p = 0 \end{align*}
Solving for \(y\) from the above results in
\begin{align*}
\tag{1} y &= \frac {\left (6 p -3\right ) x}{-1+2 p}+\frac {-3 p -1}{-1+2 p} \\
\end{align*}
This has the form
\begin{align*} y=x f(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.
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 ) = 3\\ p \left (x \right ) = {\frac {1}{2}} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
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 ) = -2\\ u \left (x \right ) = 0 \end{align*}
Substituting \(u \left (x \right )=\frac {1}{-3 x +y \left (x \right )+\frac {3}{2}}\) in the above solution gives
\[
\frac {6 x}{5}-\frac {2 y \left (x \right )}{5}-\frac {3}{5}-\frac {\ln \left (\frac {1}{-3 x +y \left (x \right )+\frac {3}{2}}\right )}{5}+\frac {\ln \left (\frac {1}{-3 x +y \left (x \right )+\frac {3}{2}}+2\right )}{5} = x +c_1
\]
Now we transform the solution \(u \left (x \right ) = -2\) to \(y \left (x \right )\) using \(u \left (x \right )=\frac {1}{-3 x +y \left (x \right )+\frac {3}{2}}\) which gives
\[
y \left (x \right ) = -2+3 x
\]
Simplifying the above gives
\begin{align*}
\frac {6 x}{5}-\frac {2 y \left (x \right )}{5}-\frac {3}{5}+\frac {\ln \left (2\right )}{5}-\frac {\ln \left (\frac {1}{-6 x +2 y \left (x \right )+3}\right )}{5}+\frac {\ln \left (\frac {y \left (x \right )-3 x +2}{-6 x +2 y \left (x \right )+3}\right )}{5} &= x +c_1 \\
y \left (x \right ) &= -2+3 x \\
\end{align*}
Solving for \(y \left (x \right )\) gives
\begin{align*}
y \left (x \right ) &= -2+3 x \\
y \left (x \right ) &= 3 x -2-\frac {\operatorname {LambertW}\left (-{\mathrm e}^{5 c_1 +5 x -1}\right )}{2} \\
\end{align*}
Figure 2.18: Slope field \(3 x -y \left (x \right )+1-\left (6 x -2 y \left (x \right )-3\right ) y^{\prime }\left (x \right ) = 0\)
Figure 2.19: Slope field \(3 x -y \left (x \right )+1-\left (6 x -2 y \left (x \right )-3\right ) y^{\prime }\left (x \right ) = 0\)
Summary of solutions found
\begin{align*}
y \left (x \right ) &= -2+3 x \\
y \left (x \right ) &= 3 x -2-\frac {\operatorname {LambertW}\left (-{\mathrm e}^{5 c_1 +5 x -1}\right )}{2} \\
\end{align*}
Solved using first_order_ode_LIE
Time used: 0.214 (sec)
Solve
\begin{align*}
3 x -y \left (x \right )+1-\left (6 x -2 y \left (x \right )-3\right ) y^{\prime }\left (x \right )&=0 \\
\end{align*}
Writing the ode as
\begin{align*} y^{\prime }\left (x \right )&=\frac {-3 x +y -1}{-6 x +2 y +3}\\ y^{\prime }\left (x \right )&= \omega \left ( x,y \left (x \right )\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
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 (-3 x +y -1\right ) B}{\left (-6 x +2 y +3\right ) C x}+\frac {3}{-6 x +2 y +3}-\frac {6 \left (-3 x +y -1\right )}{\left (-6 x +2 y +3\right )^{2}}-\frac {\left (\frac {1}{-6 x +2 y +3}-\frac {2 \left (-3 x +y -1\right )}{\left (-6 x +2 y +3\right )^{2}}\right ) \left (A x +B y \right )}{C x} = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {-18 B \,x^{3}+48 B \,x^{2} y -26 B x \,y^{2}+4 B \,y^{3}+5 A \,x^{2}+3 B \,x^{2}-32 B x y +12 B \,y^{2}-15 C \,x^{2}+3 B x +9 B y}{C \,x^{2} \left (6 x -2 y -3\right )^{2}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} 18 B \,x^{3}-48 B \,x^{2} y +26 B x \,y^{2}-4 B \,y^{3}-5 A \,x^{2}-3 B \,x^{2}+32 B x y -12 B \,y^{2}+15 C \,x^{2}-3 B x -9 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} 18 B v_{1}^{3}-48 B v_{1}^{2} v_{2}+26 B v_{1} v_{2}^{2}-4 B v_{2}^{3}-5 A v_{1}^{2}-3 B v_{1}^{2}+32 B v_{1} v_{2}-12 B v_{2}^{2}+15 C v_{1}^{2}-3 B v_{1}-9 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} 18 B v_{1}^{3}-48 B v_{1}^{2} v_{2}+\left (-5 A -3 B +15 C \right ) v_{1}^{2}+26 B v_{1} v_{2}^{2}+32 B v_{1} v_{2}-3 B v_{1}-4 B v_{2}^{3}-12 B v_{2}^{2}-9 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 {-6 x +2 y +3}{15 x -5 y -10}\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 {2 R +3}{-5 R -10} \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 \(f(R)\).
\[
y = -\frac {\operatorname {LambertW}\left (-2 \,{\mathrm e}^{5 x -4-5 c_1}\right )}{2}+3 x -2
\]
Maple trace
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) = 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
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x -y \left (x \right )+1-\left (6 x -2 y \left (x \right )-3\right ) \left (\frac {d}{d x}y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {-3 x +y \left (x \right )-1}{6 x -2 y \left (x \right )-3} \end {array} \]
✓ Mathematica. Time used: 2.097 (sec). Leaf size: 35