Internal
problem
ID
[3850] Book
:
Differential
equations
for
engineers
by
Wei-Chau
XIE,
Cambridge
Press
2010 Section
:
Chapter
2.
First-Order
and
Simple
Higher-Order
Differential
Equations.
Page
78 Problem
number
:
56 Date
solved
:
Thursday, October 17, 2024 at 04:54:00 AM CAS
classification
:
[_separable]
Solve
\begin{align*} 1-\left (y-2 x y\right ) y^{\prime }&=0 \end{align*}
1.55.1 Solved as first order separable ode
Time used: 0.178 (sec)
The ode \(y^{\prime } = -\frac {1}{y \left (2 x -1\right )}\) is separable as it can be written as
\begin{align*} f_0 &=0\\ f_1 &=-\frac {1}{2 x -1} \end{align*}
The first step is to divide the above equation by \(y^n \) which gives
\[ \frac {y'}{y^n} = f_1(x) \tag {3} \]
The next step is use the
substitution \(v = y^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be
easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is
what we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows
that
\begin{align*} f_0(x)&=0\\ f_1(x)&=-\frac {1}{2 x -1}\\ n &=-1 \end{align*}
Dividing both sides of ODE (1) by \(y^n=\frac {1}{y}\) gives
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
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (2 x y -y\right )\\ &= 2 y \end{align*}
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
\begin{align*} \mu &= e^{-\ln \left (2 x -1\right ) } \\ &= \frac {1}{2 x -1} \end{align*}
\(M\) 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 \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{2 x -1}\left (1\right ) \\ &= \frac {1}{2 x -1} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{2 x -1}\left (2 x y -y\right ) \\ &= y \end{align*}
Now a modified ODE is ontained from the original ODE, which is exact and can be solved.
The modified ODE is
Where \(f(y)\) 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 \(y\) gives
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
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}-\frac {b_{3}-a_{2}}{y \left (2 x -1\right )}-\frac {a_{3}}{y^{2} \left (2 x -1\right )^{2}}-\frac {2 \left (x a_{2}+y a_{3}+a_{1}\right )}{y \left (2 x -1\right )^{2}}-\frac {x b_{2}+y b_{3}+b_{1}}{y^{2} \left (2 x -1\right )} = 0
\end{equation}
Putting the above in normal form gives
\[
\frac {4 x^{2} y^{2} b_{2}-4 x \,y^{2} b_{2}-2 x^{2} b_{2}-4 x y b_{3}-2 y^{2} a_{3}+b_{2} y^{2}-2 x b_{1}+x b_{2}-2 y a_{1}-y a_{2}+2 y b_{3}-a_{3}+b_{1}}{y^{2} \left (2 x -1\right )^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} 4 x^{2} y^{2} b_{2}-4 x \,y^{2} b_{2}-2 x^{2} b_{2}-4 x y b_{3}-2 y^{2} a_{3}+b_{2} y^{2}-2 x b_{1}+x b_{2}-2 y a_{1}-y a_{2}+2 y b_{3}-a_{3}+b_{1} = 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}{\frac {1-2 x}{2 x y -y}}} dy \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= \frac {1}{2 x -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 {1}{2 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 {1}{2 R -1}\, dR}\\ S \left (R \right ) &= \frac {\ln \left (2 R -1\right )}{2} + c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in