Internal
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[8143] Book
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miscellaneous
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section
1.0 Problem
number
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5 Date
solved
:
Sunday, November 10, 2024 at 08:06:12 PM CAS
classification
:
[_quadrature]
Solve
\begin{align*} y^{\prime }&=y+1 \end{align*}
Solved as first order autonomous ode
Time used: 0.105 (sec)
Integrating gives
\begin{align*} \int \frac {1}{y +1}d y &= dx\\ \ln \left (y +1\right )&= x +c_1 \end{align*}
for \(y\). 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*} y = -1 \end{align*}
The following diagram is the phase line diagram. It classifies each of the above
equilibrium points as stable or not stable or semi-stable.
Solving for \(y\) gives
\begin{align*}
y &= c_1 \,{\mathrm e}^{x}-1 \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= c_1 \,{\mathrm e}^{x}-1 \\
y &= -1 \\
\end{align*}
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 \(\phi \)
\[
\phi = {\mathrm e}^{-x} y +{\mathrm e}^{-x}+ c_1
\]
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
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}{y +1}} dy \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} &= 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 {1\, dR}\\ S \left (R \right ) &= R + c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} \ln \left (y+1\right ) = x +c_2 \end{align*}
Which gives
\begin{align*} y = {\mathrm e}^{x +c_2}-1 \end{align*}
The following diagram shows solution curves of the original ode and how they transform in
the canonical coordinates space using the mapping shown.