2.1.19.3 Solved using first_order_ode_LIE

Entering first order ode LIE solver

\[\begin {aligned} x^{\prime }+2 x&={\mathrm e}^{t} \end {aligned}\]

Writing the ode as

\begin{align*} x^{\prime }&=-2 x +{\mathrm e}^{t}\\ x^{\prime }&= \omega \left ( t,x\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{t}+\omega \left ( \eta _{x}-\xi _{t}\right ) -\omega ^{2}\xi _{x}-\omega _{t}\xi -\omega _{x}\eta =0\tag {A} \end{align*}

\(\xi ,\eta \) are determined by solving (A) using ansatz. The following anstaz will now be tried.

\begin{align*} \xi &= 1\tag {1E}\\ \eta &= A t +B x\tag {2E} \end{align*}

Where the unknown coefficients are

\begin{align*} \{A, B\} \end{align*}

Substituting (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} A +\left (-2 x +{\mathrm e}^{t}\right ) B -{\mathrm e}^{t}+2 A t +2 B x = 0 \end{equation}

Putting the above in normal form gives

\[ 2 A t +B \,{\mathrm e}^{t}+A -{\mathrm e}^{t} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} 2 A t +B \,{\mathrm e}^{t}+A -{\mathrm e}^{t} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{t, x\}\) in them.

\begin{align*} \{t, {\mathrm e}^{t}\} \end{align*}

The following substitution is made in order to be able to collect on all terms with \(\{t, x\}\) in them

\[ \{t = v_{1}, {\mathrm e}^{t} = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} 2 A v_{1}+B v_{2}+A -v_{2} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\begin{align*} \{v_{1}, v_{2}\} \end{align*}

Equation (7E) now becomes

\begin{equation} \tag{8E} 2 A v_{1}+\left (B -1\right ) v_{2}+A = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} A&=0 \\ 2 A&=0 \\ B -1&=0 \\ \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} A&=0\\ B&=1 \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= 1 \\ \eta &= x \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( t,x\right ) \to \left ( R,S \right )\) 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

\begin{align*} \frac {d t}{\xi } &= \frac {d x}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial t} + \eta \frac {\partial }{\partial x}\right ) S(t,x) = 1\). 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 \(S(R)\). Therefore

\begin{align*} \frac {dx}{dt} &= \frac {\eta }{\xi }\\ &= \frac {x}{1}\\ &= x \end{align*}

The solution to the above ode is

\begin{align*} x = c_1 \,{\mathrm e}^{t} \end{align*}

The coordinate \(R\) is taken as the constant of integration. Therefore

\begin{align*} R &= {\mathrm e}^{-t} x \end{align*}

And \(S\) is found from

\begin{align*} dS &= \frac {dt}{\xi } \\ &= \frac {dt}{1} \end{align*}

Integrating gives

\begin{align*} S &= \int { \frac {dt}{1}}\\ &= t \end{align*}

Where the constant of integration is set to zero as any particular solution will work. Now that \(R,S\) are found, the ode is now setupin these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{t} + \omega (t,x) S_{x} }{ R_{t} + \omega (t,x) R_{x} }\tag {2} \end{align*}

Where in the above \(R_{t},R_{x},S_{t},S_{x}\) are all partial derivatives and \(\omega (t,x)\) is the right hand side of the original ode given by

\begin{align*} \omega (t,x) &= -2 x +{\mathrm e}^{t} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{t} &= -{\mathrm e}^{-t} x\\ R_{x} &= {\mathrm e}^{-t}\\ S_{t} &= 1\\ S_{x} &= 0 \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {1}{1-3 \,{\mathrm e}^{-t} x}\tag {2A} \end{align*}

To express the RHS as function of \(R\) only, \(t,x\) are expressed in terms of \(R,S\) from the result obtained earlier and the result is simplified. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {1}{-1+3 R} \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 \(R,S\).

Because the ODE has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), the solution requires only integration. Therefore

\begin{align*} dS &= \left (-\frac {1}{-1+3 R}\right ) \, dR\\ S &= \int { \left (-\frac {1}{-1+3 R}\right ) \, dR}\\ &= -\frac {\ln \left (-1+3 R \right )}{3}+c_2 \end{align*}

To complete the solution, The above is transformed back to the \(t,x\) coordinates. This results in

\begin{align*} t = -\frac {\ln \left (3 \,{\mathrm e}^{-t} x-1\right )}{3}+c_2 \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.

Original ode in \(t,x\) coordinates	Canonical coordinates transformation	ODE in canonical coordinates \((R,S)\)
\( \frac {dx}{dt} = -2 x +{\mathrm e}^{t}\)		\( \frac {d S}{d R} = -\frac {1}{-1+3 R}\)
	\(\!\begin {aligned} R&= {\mathrm e}^{-t} x\\ S&= t \end {aligned} \)

Solving for \(x\) from \(t = -\frac {\ln \left (3 \,{\mathrm e}^{-t} x-1\right )}{3}+c_2\) gives

\begin{align*} x &= \frac {{\mathrm e}^{-3 t +3 c_2} {\mathrm e}^{t}}{3}+\frac {{\mathrm e}^{t}}{3} \\ \end{align*}

Direction field with Isoclines	Direction field

Summary of solutions found

\[ x = \frac {{\mathrm e}^{-3 t +3 c_2} {\mathrm e}^{t}}{3}+\frac {{\mathrm e}^{t}}{3} \]