Internal
problem
ID
[1662] 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
:
35(a) Date
solved
:
Thursday, October 17, 2024 at 11:42:44 AM CAS
classification
:
[[_homogeneous, `class A`], _rational, _Riccati]
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 \(M=-4 x^{2}+x y +y^{2}\) and \(N=x^{2}\) are both homogeneous and of the same order \(n=2\). 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
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 \(u^{2}-4=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=-2\\ u \left (x \right )&=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.
Therefore the solutions found are
\begin{align*} \ln \left (\frac {\left (u \left (x \right )-2\right )^{{1}/{4}}}{\left (u \left (x \right )+2\right )^{{1}/{4}}}\right ) = \ln \left (x \right )+c_1\\ u \left (x \right ) = -2\\ u \left (x \right ) = 2 \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&=-2\\ u \left (x \right )&=2\\ u \left (x \right )&=-\frac {2 \left (x^{4} {\mathrm e}^{4 c_1}+1\right )}{x^{4} {\mathrm e}^{4 c_1}-1} \end{align*}
Converting \(u \left (x \right ) = -2\) back to \(y\) gives
\begin{align*} y = -2 x \end{align*}
Converting \(u \left (x \right ) = 2\) back to \(y\) gives
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 \(u^{2}-4=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=-2\\ u \left (x \right )&=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.
Therefore the solutions found are
\begin{align*} \ln \left (\frac {\left (u \left (x \right )-2\right )^{{1}/{4}}}{\left (u \left (x \right )+2\right )^{{1}/{4}}}\right ) = \ln \left (x \right )+c_1\\ u \left (x \right ) = -2\\ u \left (x \right ) = 2 \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&=-2\\ u \left (x \right )&=2\\ u \left (x \right )&=-\frac {2 \left (x^{4} {\mathrm e}^{4 c_1}+1\right )}{x^{4} {\mathrm e}^{4 c_1}-1} \end{align*}
Converting \(u \left (x \right ) = -2\) back to \(y\) gives
\begin{align*} y = -2 x \end{align*}
Converting \(u \left (x \right ) = 2\) back to \(y\) gives
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 \(M=-4 X^{2}+X Y +Y^{2}\) and \(N=X^{2}\) are both homogeneous and of the same order \(n=2\). 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
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 \(u^{2}-4=0\) for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=-2\\ u \left (X \right )&=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.
Therefore the solutions found are
\begin{align*} \ln \left (\frac {\left (u \left (X \right )-2\right )^{{1}/{4}}}{\left (u \left (X \right )+2\right )^{{1}/{4}}}\right ) = \ln \left (X \right )+c_1\\ u \left (X \right ) = -2\\ u \left (X \right ) = 2 \end{align*}
Solving for \(u \left (X \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (X \right )&=-2\\ u \left (X \right )&=2\\ u \left (X \right )&=-\frac {2 \left (X^{4} {\mathrm e}^{4 c_1}+1\right )}{X^{4} {\mathrm e}^{4 c_1}-1} \end{align*}
Converting \(u \left (X \right ) = -2\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = -2 X \end{align*}
Converting \(u \left (X \right ) = 2\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = 2 X \end{align*}
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 \(u^{2}-4=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=-2\\ u \left (x \right )&=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.
Therefore the solutions found are
\begin{align*} \ln \left (\frac {\left (u \left (x \right )-2\right )^{{1}/{4}}}{\left (u \left (x \right )+2\right )^{{1}/{4}}}\right ) = \ln \left (x \right )+c_1\\ u \left (x \right ) = -2\\ u \left (x \right ) = 2 \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&=-2\\ u \left (x \right )&=2\\ u \left (x \right )&=-\frac {2 \left (x^{4} {\mathrm e}^{4 c_1}+1\right )}{x^{4} {\mathrm e}^{4 c_1}-1} \end{align*}
Converting \(u \left (x \right ) = -2\) back to \(y\) gives
\begin{align*} \frac {y}{x} = -2 \end{align*}
Converting \(u \left (x \right ) = 2\) back to \(y\) gives
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 {4 x^{2}-y^{2}}{x}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (2 x +y \right )}{4}-\frac {\ln \left (y -2 x \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 \(f(R)\).
This is Euler second order ODE. Let the solution be \(u = x^r\), then \(u'=r x^{r-1}\) and \(u''=r(r-1) x^{r-2}\). Substituting these back
into the given ODE gives
\[ x^{2}(r(r-1))x^{r-2}+x r x^{r-1}-4 x^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )x^{r}+r\,x^{r}-4 x^{r} = 0 \]
Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives
\[ r \left (r -1\right )+r-4 = 0 \]
Or
\[ r^{2}-4 = 0 \tag {1} \]
Equation (1) is the characteristic equation. Its roots determine the form of the general
solution. Using the quadratic equation the roots are
\begin{align*} r_1 &= -2\\ r_2 &= 2 \end{align*}
Since the roots are real and distinct, then the general solution is
\[ u= c_1 u_1 + c_2 u_2 \]
Where \(u_1 = x^{r_1}\) and \(u_2 = x^{r_2} \). Hence
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous D<-homogeneous successful`