5.21 Internal
problem
ID
[7962]Book
:
Own
collection
of
miscellaneous
problems Section
:
section
5.0 Problem
number
:
21Date
solved
:
Monday, October 21, 2024 at 04:39:40 PMCAS
classification
:
[_separable]
Solve
\begin{align*} x^{2} y^{\prime }+{\mathrm e}^{-y}&=0 \end{align*}
5.21.1 Time used: 0.106 (sec)
The ode \(y^{\prime } = -\frac {{\mathrm e}^{-y}}{x^{2}}\) is separable as it can be written as
\begin{align*} y^{\prime }&= -\frac {{\mathrm e}^{-y}}{x^{2}}\\ &= f(x) g(y) \end{align*}
Where
\begin{align*} f(x) &= -\frac {1}{x^{2}}\\ g(y) &= {\mathrm e}^{-y} \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { {\mathrm e}^{y}\,dy} &= \int { -\frac {1}{x^{2}} \,dx}\\ {\mathrm e}^{y}&=\frac {1}{x}+c_1 \end{align*}
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} y = \ln \left (\frac {c_1 x +1}{x}\right ) \end{align*}
Figure 219: Slope field plot\(x^{2} y^{\prime }+{\mathrm e}^{-y} = 0\) 5.21.2 Time used: 0.560 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=-\frac {{\mathrm e}^{-y}}{x^{2}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to
use as anstaz gives
\begin{align*}
\tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\
\tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\
\end{align*}
Where the unknown coefficients are
\[
\{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\}
\]
Substituting equations (1E,2E) and \(\omega \)
into (A) gives
\begin{equation}
\tag{5E} b_{2}-\frac {{\mathrm e}^{-y} \left (b_{3}-a_{2}\right )}{x^{2}}-\frac {{\mathrm e}^{-2 y} a_{3}}{x^{4}}-\frac {2 \,{\mathrm e}^{-y} \left (x a_{2}+y a_{3}+a_{1}\right )}{x^{3}}-\frac {{\mathrm e}^{-y} \left (x b_{2}+y b_{3}+b_{1}\right )}{x^{2}} = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {{\mathrm e}^{-y} x^{3} b_{2}+{\mathrm e}^{-y} x^{2} y b_{3}-b_{2} x^{4}+{\mathrm e}^{-y} x^{2} a_{2}+{\mathrm e}^{-y} x^{2} b_{1}+{\mathrm e}^{-y} x^{2} b_{3}+2 \,{\mathrm e}^{-y} x y a_{3}+{\mathrm e}^{-2 y} a_{3}+2 \,{\mathrm e}^{-y} x a_{1}}{x^{4}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -{\mathrm e}^{-y} x^{3} b_{2}-{\mathrm e}^{-y} x^{2} y b_{3}+b_{2} x^{4}-{\mathrm e}^{-y} x^{2} a_{2}-{\mathrm e}^{-y} x^{2} b_{1}-{\mathrm e}^{-y} x^{2} b_{3}-2 \,{\mathrm e}^{-y} x y a_{3}-{\mathrm e}^{-2 y} a_{3}-2 \,{\mathrm e}^{-y} x a_{1} = 0
\end{equation}
Simplifying the above gives
\begin{equation}
\tag{6E} -{\mathrm e}^{-y} x^{3} b_{2}-{\mathrm e}^{-y} x^{2} y b_{3}+b_{2} x^{4}-{\mathrm e}^{-y} x^{2} a_{2}-{\mathrm e}^{-y} x^{2} b_{1}-{\mathrm e}^{-y} x^{2} b_{3}-2 \,{\mathrm e}^{-y} x y a_{3}-{\mathrm e}^{-2 y} a_{3}-2 \,{\mathrm e}^{-y} x a_{1} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y, {\mathrm e}^{-2 y}, {\mathrm e}^{-y}\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\{x = v_{1}, y = v_{2}, {\mathrm e}^{-2 y} = v_{3}, {\mathrm e}^{-y} = v_{4}\}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} b_{2} v_{1}^{4}-v_{4} v_{1}^{3} b_{2}-v_{4} v_{1}^{2} v_{2} b_{3}-v_{4} v_{1}^{2} a_{2}-2 v_{4} v_{1} v_{2} a_{3}-v_{4} v_{1}^{2} b_{1}-v_{4} v_{1}^{2} b_{3}-2 v_{4} v_{1} a_{1}-v_{3} a_{3} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}, v_{4}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} b_{2} v_{1}^{4}-v_{4} v_{1}^{3} b_{2}-v_{4} v_{1}^{2} v_{2} b_{3}+\left (-a_{2}-b_{1}-b_{3}\right ) v_{1}^{2} v_{4}-2 v_{4} v_{1} v_{2} a_{3}-2 v_{4} v_{1} a_{1}-v_{3} a_{3} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} b_{2}&=0\\ -2 a_{1}&=0\\ -2 a_{3}&=0\\ -a_{3}&=0\\ -b_{2}&=0\\ -b_{3}&=0\\ -a_{2}-b_{1}-b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=-b_{1}\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=0 \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 &= -x \\
\eta &= 1 \\
\end{align*}
Shifting is now applied to make \(\xi =0\) in order to simplify the rest of
the computation
\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= 1 - \left (-\frac {{\mathrm e}^{-y}}{x^{2}}\right ) \left (-x\right ) \\ &= \frac {-{\mathrm e}^{-y}+x}{x}\\ \xi &= 0 \end{align*}
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
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
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 {-{\mathrm e}^{-y}+x}{x}}} dy \end{align*}
Which results in
\begin{align*} S&= \ln \left (x \,{\mathrm e}^{y}-1\right ) \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}
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
\begin{align*} \omega (x,y) &= -\frac {{\mathrm e}^{-y}}{x^{2}} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {{\mathrm e}^{y}}{x \,{\mathrm e}^{y}-1}\\ S_{y} &= \frac {x \,{\mathrm e}^{y}}{x \,{\mathrm e}^{y}-1} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= \frac {1}{x}\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}{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 \(R,S\) .
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}{R}\, dR}\\ S \left (R \right ) &= \ln \left (R \right ) + 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 (x \,{\mathrm e}^{y}-1\right ) = \ln \left (x \right )+c_2 \end{align*}
Which gives
\begin{align*} y = \ln \left (\frac {1+x \,{\mathrm e}^{c_2}}{x}\right ) \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 \(x,y\) coordinates
Canonical
coordinates
transformation
ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = -\frac {{\mathrm e}^{-y}}{x^{2}}\)
\( \frac {d S}{d R} = \frac {1}{R}\)
\(\!\begin {aligned} R&= x\\ S&= \ln \left (x \,{\mathrm e}^{y}-1\right ) \end {aligned} \)
Figure 220: Slope field plot\(x^{2} y^{\prime }+{\mathrm e}^{-y} = 0\) 5.21.3 Time used: 0.111 (sec)
Writing the ode as
\begin{align*} y^{\prime } &= -\frac {{\mathrm e}^{-y}}{x^{2}}\tag {1} \end{align*}
And using the substitution \(u={\mathrm e}^{y}\) then
\begin{align*} u' &= y^{\prime } {\mathrm e}^{y} \end{align*}
The above shows that
\begin{align*} y^{\prime } &= u^{\prime }\left (x \right ) {\mathrm e}^{-y}\\ &= \frac {u^{\prime }\left (x \right )}{u} \end{align*}
Substituting this in (1) gives
\begin{align*} \frac {u^{\prime }\left (x \right )}{u}&=-\frac {1}{x^{2} u} \end{align*}
The above simplifies to
\begin{align*} u^{\prime }\left (x \right )&=-\frac {1}{x^{2}}\tag {2} \end{align*}
Now ode (2) is solved for \(u \left (x \right )\) .
Since the ode has the form \(u^{\prime }\left (x \right )=f(x)\) , then we only need to integrate \(f(x)\) .
\begin{align*} \int {du} &= \int {-\frac {1}{x^{2}}\, dx}\\ u \left (x \right ) &= \frac {1}{x} + c_1 \end{align*}
Substituting the solution found for \(u \left (x \right )\) in \(u={\mathrm e}^{y}\) gives
\begin{align*} y&= \ln \left (u \left (x \right )\right )\\ &= \ln \left (\ln \left (\frac {1}{x}+c_1 \right )\right )\\ &= \ln \left (\frac {1}{x}+c_1 \right ) \end{align*}
Figure 221: Slope field plot\(x^{2} y^{\prime }+{\mathrm e}^{-y} = 0\) 5.21.4 \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }+{\mathrm e}^{-y}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{-y}}{x^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\mathrm e}^{-y}}=-\frac {1}{x^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\mathrm e}^{-y}}d x =\int -\frac {1}{x^{2}}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {1}{{\mathrm e}^{-y}}=\frac {1}{x}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\ln \left (\frac {x}{\mathit {C1} x +1}\right ) \end {array} \]
5.21.5 Methods for first order ODEs:
5.21.6 Solving time : 0.004
dsolve ( x ^2* diff ( y ( x ), x )+ exp (- y ( x )) = 0,
y(x),singsol=all)
\[
y = \ln \left (\frac {-c_1 x +1}{x}\right )
\]
5.21.7 Solving time : 0.481
DSolve [{ x ^2* D [ y [ x ], x ]+ Exp [- y [ x ]]==0,{}}, y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to \log \left (\frac {1}{x}+c_1\right )
\]