Internal
problem
ID
[8146]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
8
Date
solved
:
Sunday, November 10, 2024 at 08:06:13 PM
CAS
classification
:
[_quadrature]
Solve
Time used: 0.083 (sec)
Integrating gives
The following diagram is the phase line diagram. It classifies each of the above equilibrium points as stable or not stable or semi-stable.
Summary of solutions found
Time used: 0.135 (sec)
Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes
Which is now solved The ode \(u^{\prime }\left (x \right ) = \frac {u \left (x \right ) \left (x -1\right )}{x}\) is separable as it can be written as
Where
Integrating 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 \(u=0\) for \(u \left (x \right )\) gives
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
Solving for \(u \left (x \right )\) gives
Converting \(u \left (x \right ) = 0\) back to \(y\) gives
Converting \(u \left (x \right ) = \frac {{\mathrm e}^{x +c_1}}{x}\) back to \(y\) gives
Summary of solutions found
Time used: 0.098 (sec)
To solve an ode of the form
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
Hence
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
Therefore
Comparing (1A) and (2A) shows that
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied
Using result found above gives
And
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
Since \(A\) does not depend on \(y\), then it can be used to find an integrating factor. The integrating factor \(\mu \) is
The result of integrating gives
\(M\) 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 \(N\).
And
Now a modified ODE is ontained from the original ODE, which is exact and can be solved. The modified ODE is
The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)
Integrating (2) w.r.t. \(y\) gives
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
But equation (1) says that \(\frac {\partial \phi }{\partial x} = -y \,{\mathrm e}^{-x}\). Therefore equation (4) becomes
Solving equation (5) for \( f'(x)\) gives
Therefore
Where \(c_1\) is constant of integration. Substituting this result for \(f(x)\) into equation (3) gives \(\phi \)
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
Solving for \(y\) gives
Summary of solutions found
Time used: 0.363 (sec)
Writing the ode as
The condition of Lie symmetry is the linearized PDE given by
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives
Where the unknown coefficients are
Substituting equations (1E,2E) and \(\omega \) into (A) gives
Putting the above in normal form gives
Setting the numerator to zero gives
Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.
The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them
The above PDE (6E) now becomes
Collecting the above on the terms \(v_i\) introduced, and these are
Equation (7E) now becomes
Setting each coefficients in (8E) to zero gives the following equations to solve
Solving the above equations for the unknowns gives
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) 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
\(S\) is found from
Which results in
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by
Evaluating all the partial derivatives gives
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
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 \(f(R)\).
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in
Which gives
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 \((R,S)\) |
\( \frac {dy}{dx} = y\) |
|
\( \frac {d S}{d R} = 1\) |
|
\(\!\begin {aligned} R&= x\\ S&= \ln \left (y \right ) \end {aligned} \) |
|
Summary of solutions found
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
Solving time : 0.001
(sec)
Leaf size : 8
dsolve(diff(y(x),x) = y(x), y(x),singsol=all)