Internal
problem
ID
[14769] Book
:
INTRODUCTORY
DIFFERENTIAL
EQUATIONS.
Martha
L.
Abell,
James
P.
Braselton.
Fourth
edition
2014.
ElScAe.
2014 Section
:
Chapter
1.
Introduction
to
Differential
Equations.
Review
exercises,
page
23 Problem
number
:
20 Date
solved
:
Friday, October 11, 2024 at 05:11:57 PM CAS
classification
:
[[_linear, `class A`]]
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\)
for now so not to confuse them with the original \(M\) and \(N\).
Where \(f(x)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(x\) gives
Where \(c_1\) is constant of integration. Substituting result found above for \(f(x)\) into
equation (3) gives \(\phi \)
\[
\phi = y \,{\mathrm e}^{2 x}-\frac {\left (2 x^{2}-2 x +1\right ) {\mathrm e}^{2 x}}{4}+ c_1
\]
But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new
constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[
c_1 = y \,{\mathrm e}^{2 x}-\frac {\left (2 x^{2}-2 x +1\right ) {\mathrm e}^{2 x}}{4}
\]
Solving for the constant of integration from initial conditions, the solution becomes
\begin{align*} y \,{\mathrm e}^{2 x}-\frac {\left (2 x^{2}-2 x +1\right ) {\mathrm e}^{2 x}}{4} = {\frac {3}{4}} \end{align*}
2.15.4 Solved using Lie symmetry for first order ode
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}+\left (x^{2}-2 y \right ) \left (b_{3}-a_{2}\right )-\left (x^{2}-2 y \right )^{2} a_{3}-2 x \left (x a_{2}+y a_{3}+a_{1}\right )+2 x b_{2}+2 y b_{3}+2 b_{1} = 0
\end{equation}
Putting the above in normal form gives
\[
-x^{4} a_{3}+4 x^{2} y a_{3}-3 x^{2} a_{2}+x^{2} b_{3}-2 x y a_{3}-4 y^{2} a_{3}-2 x a_{1}+2 x b_{2}+2 y a_{2}+2 b_{1}+b_{2} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -x^{4} a_{3}+4 x^{2} y a_{3}-3 x^{2} a_{2}+x^{2} b_{3}-2 x y a_{3}-4 y^{2} a_{3}-2 x a_{1}+2 x b_{2}+2 y a_{2}+2 b_{1}+b_{2} = 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
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}{2 x^{2}-2 x -4 y +1}} dy \end{align*}
Which results in
\begin{align*} S&= -\frac {\ln \left (2 x^{2}-2 x -4 y +1\right )}{4} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
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
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 \(f(R)\).