Internal problem ID [2029]
Internal file name [OUTPUT/2029_Sunday_February_25_2024_06_45_33_AM_35347104/index.tex
]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 11, page 45
Problem number: 19.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(y)]`]]
\[ \boxed {y^{\prime }-x \left (1-{\mathrm e}^{2 y-x^{2}}\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -x \left ({\mathrm e}^{-x^{2}+2 y}-1\right ) \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=0\) is \[
\{-\infty The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=0\) is \[
\{-\infty
Writing the ode as \begin {align*} y^{\prime }&=-x \left ({\mathrm e}^{-x^{2}+2 y}-1\right )\\ 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*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using
ansatz. Making bivariate polynomials of degree 2 to use as anstaz gives \begin{align*}
\tag{1E} \xi &= x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\
\tag{2E} \eta &= x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\
\end{align*} Where
the unknown coefficients are \[
\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\}
\] Substituting equations (1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} 2 x b_{4}+y b_{5}+b_{2}-x \left ({\mathrm e}^{-x^{2}+2 y}-1\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )-x^{2} \left ({\mathrm e}^{-x^{2}+2 y}-1\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )-\left (-{\mathrm e}^{-x^{2}+2 y}+1+2 x^{2} {\mathrm e}^{-x^{2}+2 y}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )+2 x \,{\mathrm e}^{-x^{2}+2 y} \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation} Putting the above in normal form gives \[
-2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} y a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y^{2} a_{6}+4 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y a_{6}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y b_{5}+2 \,{\mathrm e}^{-x^{2}+2 y} x \,y^{2} b_{6}+2 \,{\mathrm e}^{-x^{2}+2 y} x y a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x y b_{6}+2 x b_{4}+y b_{5}+x^{2} b_{5}-x^{3} a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} a_{2}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{1}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{3}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} b_{2}+2 \,{\mathrm e}^{-x^{2}+2 y} x a_{2}+2 \,{\mathrm e}^{-x^{2}+2 y} x b_{1}-{\mathrm e}^{-x^{2}+2 y} x b_{3}+{\mathrm e}^{-x^{2}+2 y} y a_{3}-x^{2} a_{3}-{\mathrm e}^{-2 x^{2}+4 y} x^{2} a_{3}-2 \,{\mathrm e}^{-2 x^{2}+4 y} x^{2} y a_{6}-3 x^{2} a_{4}-y^{2} a_{6}-2 y x a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y a_{3}+2 \,{\mathrm e}^{-x^{2}+2 y} x y b_{3}+2 x y b_{6}-2 x^{2} y a_{6}-a_{1}+b_{2}+x b_{3}+{\mathrm e}^{-x^{2}+2 y} a_{1}-{\mathrm e}^{-2 x^{2}+4 y} x^{3} a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{4} a_{4}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} a_{5}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} b_{4}+3 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{4}-{\mathrm e}^{-x^{2}+2 y} x^{2} b_{5}+{\mathrm e}^{-x^{2}+2 y} y^{2} a_{6}-2 x a_{2}-y a_{3} = 0
\] Setting the numerator to zero gives \begin{equation}
\tag{6E} -2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} y a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y^{2} a_{6}+4 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y a_{6}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y b_{5}+2 \,{\mathrm e}^{-x^{2}+2 y} x \,y^{2} b_{6}+2 \,{\mathrm e}^{-x^{2}+2 y} x y a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x y b_{6}+2 x b_{4}+y b_{5}+x^{2} b_{5}-x^{3} a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} a_{2}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{1}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{3}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} b_{2}+2 \,{\mathrm e}^{-x^{2}+2 y} x a_{2}+2 \,{\mathrm e}^{-x^{2}+2 y} x b_{1}-{\mathrm e}^{-x^{2}+2 y} x b_{3}+{\mathrm e}^{-x^{2}+2 y} y a_{3}-x^{2} a_{3}-{\mathrm e}^{-2 x^{2}+4 y} x^{2} a_{3}-2 \,{\mathrm e}^{-2 x^{2}+4 y} x^{2} y a_{6}-3 x^{2} a_{4}-y^{2} a_{6}-2 y x a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y a_{3}+2 \,{\mathrm e}^{-x^{2}+2 y} x y b_{3}+2 x y b_{6}-2 x^{2} y a_{6}-a_{1}+b_{2}+x b_{3}+{\mathrm e}^{-x^{2}+2 y} a_{1}-{\mathrm e}^{-2 x^{2}+4 y} x^{3} a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{4} a_{4}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} a_{5}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} b_{4}+3 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{4}-{\mathrm e}^{-x^{2}+2 y} x^{2} b_{5}+{\mathrm e}^{-x^{2}+2 y} y^{2} a_{6}-2 x a_{2}-y a_{3} = 0
\end{equation}
Simplifying the above gives \begin{equation}
\tag{6E} -2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} y a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y^{2} a_{6}+4 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y a_{6}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y b_{5}+2 \,{\mathrm e}^{-x^{2}+2 y} x \,y^{2} b_{6}+2 \,{\mathrm e}^{-x^{2}+2 y} x y a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x y b_{6}+2 x b_{4}+y b_{5}+x^{2} b_{5}-x^{3} a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} a_{2}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{1}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{3}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} b_{2}+2 \,{\mathrm e}^{-x^{2}+2 y} x a_{2}+2 \,{\mathrm e}^{-x^{2}+2 y} x b_{1}-{\mathrm e}^{-x^{2}+2 y} x b_{3}+{\mathrm e}^{-x^{2}+2 y} y a_{3}-x^{2} a_{3}-{\mathrm e}^{-2 x^{2}+4 y} x^{2} a_{3}-2 \,{\mathrm e}^{-2 x^{2}+4 y} x^{2} y a_{6}-3 x^{2} a_{4}-y^{2} a_{6}-2 y x a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{2} y a_{3}+2 \,{\mathrm e}^{-x^{2}+2 y} x y b_{3}+2 x y b_{6}-2 x^{2} y a_{6}-a_{1}+b_{2}+x b_{3}+{\mathrm e}^{-x^{2}+2 y} a_{1}-{\mathrm e}^{-2 x^{2}+4 y} x^{3} a_{5}-2 \,{\mathrm e}^{-x^{2}+2 y} x^{4} a_{4}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} a_{5}+2 \,{\mathrm e}^{-x^{2}+2 y} x^{3} b_{4}+3 \,{\mathrm e}^{-x^{2}+2 y} x^{2} a_{4}-{\mathrm e}^{-x^{2}+2 y} x^{2} b_{5}+{\mathrm e}^{-x^{2}+2 y} y^{2} a_{6}-2 x a_{2}-y a_{3} = 0
\end{equation} Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them. \[
\{x, y, {\mathrm e}^{-2 x^{2}+4 y}, {\mathrm e}^{-x^{2}+2 y}\}
\] The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them \[
\{x = v_{1}, y = v_{2}, {\mathrm e}^{-2 x^{2}+4 y} = v_{3}, {\mathrm e}^{-x^{2}+2 y} = v_{4}\}
\] The above PDE (6E) now becomes \begin{equation}
\tag{7E} -2 v_{4} v_{1}^{4} a_{4}-2 v_{4} v_{1}^{3} v_{2} a_{5}-2 v_{4} v_{1}^{2} v_{2}^{2} a_{6}-2 v_{4} v_{1}^{3} a_{2}-2 v_{4} v_{1}^{2} v_{2} a_{3}-v_{3} v_{1}^{3} a_{5}+2 v_{4} v_{1}^{3} a_{5}-2 v_{3} v_{1}^{2} v_{2} a_{6}+4 v_{4} v_{1}^{2} v_{2} a_{6}+2 v_{4} v_{1}^{3} b_{4}+2 v_{4} v_{1}^{2} v_{2} b_{5}+2 v_{4} v_{1} v_{2}^{2} b_{6}-2 v_{4} v_{1}^{2} a_{1}-v_{3} v_{1}^{2} a_{3}+2 v_{4} v_{1}^{2} a_{3}+3 v_{4} v_{1}^{2} a_{4}-v_{1}^{3} a_{5}+2 v_{4} v_{1} v_{2} a_{5}-2 v_{1}^{2} v_{2} a_{6}+v_{4} v_{2}^{2} a_{6}+2 v_{4} v_{1}^{2} b_{2}+2 v_{4} v_{1} v_{2} b_{3}-v_{4} v_{1}^{2} b_{5}-2 v_{4} v_{1} v_{2} b_{6}+2 v_{4} v_{1} a_{2}-v_{1}^{2} a_{3}+v_{4} v_{2} a_{3}-3 v_{1}^{2} a_{4}-2 v_{2} v_{1} a_{5}-v_{2}^{2} a_{6}+2 v_{4} v_{1} b_{1}-v_{4} v_{1} b_{3}+v_{1}^{2} b_{5}+2 v_{1} v_{2} b_{6}+v_{4} a_{1}-2 v_{1} a_{2}-v_{2} a_{3}+v_{1} b_{3}+2 v_{1} b_{4}+v_{2} b_{5}-a_{1}+b_{2} = 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} -2 v_{4} v_{1}^{4} a_{4}-2 v_{4} v_{1}^{3} v_{2} a_{5}-v_{3} v_{1}^{3} a_{5}+\left (-2 a_{2}+2 a_{5}+2 b_{4}\right ) v_{1}^{3} v_{4}-v_{1}^{3} a_{5}-2 v_{4} v_{1}^{2} v_{2}^{2} a_{6}-2 v_{3} v_{1}^{2} v_{2} a_{6}+\left (-2 a_{3}+4 a_{6}+2 b_{5}\right ) v_{1}^{2} v_{2} v_{4}-2 v_{1}^{2} v_{2} a_{6}-v_{3} v_{1}^{2} a_{3}+\left (-2 a_{1}+2 a_{3}+3 a_{4}+2 b_{2}-b_{5}\right ) v_{1}^{2} v_{4}+\left (-a_{3}-3 a_{4}+b_{5}\right ) v_{1}^{2}+2 v_{4} v_{1} v_{2}^{2} b_{6}+\left (2 a_{5}+2 b_{3}-2 b_{6}\right ) v_{1} v_{2} v_{4}+\left (-2 a_{5}+2 b_{6}\right ) v_{1} v_{2}+\left (2 a_{2}+2 b_{1}-b_{3}\right ) v_{1} v_{4}+\left (-2 a_{2}+b_{3}+2 b_{4}\right ) v_{1}+v_{4} v_{2}^{2} a_{6}-v_{2}^{2} a_{6}+v_{4} v_{2} a_{3}+\left (-a_{3}+b_{5}\right ) v_{2}+v_{4} a_{1}-a_{1}+b_{2} = 0
\end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve
\begin {align*} a_{1}&=0\\ a_{3}&=0\\ a_{6}&=0\\ -a_{3}&=0\\ -2 a_{4}&=0\\ -2 a_{5}&=0\\ -a_{5}&=0\\ -2 a_{6}&=0\\ -a_{6}&=0\\ 2 b_{6}&=0\\ -a_{1}+b_{2}&=0\\ -a_{3}+b_{5}&=0\\ -2 a_{5}+2 b_{6}&=0\\ -2 a_{2}+2 a_{5}+2 b_{4}&=0\\ -2 a_{2}+b_{3}+2 b_{4}&=0\\ 2 a_{2}+2 b_{1}-b_{3}&=0\\ -2 a_{3}+4 a_{6}+2 b_{5}&=0\\ -a_{3}-3 a_{4}+b_{5}&=0\\ 2 a_{5}+2 b_{3}-2 b_{6}&=0\\ -2 a_{1}+2 a_{3}+3 a_{4}+2 b_{2}-b_{5}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=b_{4}\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=-b_{4}\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=b_{4}\\ b_{5}&=0\\ b_{6}&=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 &= x^{2}-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 \\ &= x^{2}-1 - \left (-x \left ({\mathrm e}^{-x^{2}+2 y}-1\right )\right ) \left (x\right ) \\ &= x^{2} {\mathrm e}^{-x^{2}} {\mathrm e}^{2 y}-1\\ \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}{x^{2} {\mathrm e}^{-x^{2}} {\mathrm e}^{2 y}-1}} dy \end {align*}
Which results in \begin {align*} S&= \frac {{\mathrm e}^{x^{2}} {\mathrm e}^{-x^{2}} \ln \left (-x^{2} {\mathrm e}^{2 y}+{\mathrm e}^{x^{2}}\right )}{2}-\ln \left ({\mathrm e}^{y}\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) &= -x \left ({\mathrm e}^{-x^{2}+2 y}-1\right ) \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {x \left ({\mathrm e}^{x^{2}}-{\mathrm e}^{2 y}\right )}{-x^{2} {\mathrm e}^{2 y}+{\mathrm e}^{x^{2}}}\\ S_{y} &= \frac {{\mathrm e}^{x^{2}}}{x^{2} {\mathrm e}^{2 y}-{\mathrm e}^{x^{2}}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin {align*} \frac {dS}{dR} &= 0\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} &= 0 \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 \(R,S\). Integrating the above gives
\begin {align*} S \left (R \right ) = c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in
\begin {align*} \frac {\ln \left (-x^{2} {\mathrm e}^{2 y}+{\mathrm e}^{x^{2}}\right )}{2}-y = c_{1} \end {align*}
Which simplifies to \begin {align*} \frac {\ln \left (-x^{2} {\mathrm e}^{2 y}+{\mathrm e}^{x^{2}}\right )}{2}-y = c_{1} \end {align*}
Which gives \begin {align*} y = \frac {\ln \left (\frac {1}{{\mathrm e}^{2 c_{1}}+x^{2}}\right )}{2}+\frac {x^{2}}{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 \(x,y\) coordinates
Canonical
coordinates
transformation ODE in canonical coordinates \((R,S)\) \( \frac {dy}{dx} = -x \left ({\mathrm e}^{-x^{2}+2 y}-1\right )\) \( \frac {d S}{d R} = 0\) \(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (-x^{2} {\mathrm e}^{2 y}+{\mathrm e}^{x^{2}}\right )}{2}-y \end {aligned} \) Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 0 = \frac {\ln \left ({\mathrm e}^{-2 c_{1}}\right )}{2} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\ln \left (\frac {1}{x^{2}+1}\right )}{2}+\frac {x^{2}}{2} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\ln \left (\frac {1}{x^{2}+1}\right )}{2}+\frac {x^{2}}{2} \\
\end{align*} Verification of solutions
\[
y = \frac {\ln \left (\frac {1}{x^{2}+1}\right )}{2}+\frac {x^{2}}{2}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-x \left (1-{\mathrm e}^{2 y-x^{2}}\right )=0, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x \left (1-{\mathrm e}^{2 y-x^{2}}\right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.093 (sec). Leaf size: 18
\[
y \left (x \right ) = \frac {x^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}
\]
✓ Solution by Mathematica
Time used: 0.5 (sec). Leaf size: 21
\[
y(x)\to \frac {1}{2} \left (x^2-\log \left (x^2+1\right )\right )
\]
7.19.2 Solving as first order ode lie symmetry calculated ode
7.19.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
trying inverse linear
trying homogeneous types:
trying Chini
differential order: 1; looking for linear symmetries
trying exact
Looking for potential symmetries
trying inverse_Riccati
trying an equivalence to an Abel ODE
differential order: 1; trying a linearization to 2nd order
--- trying a change of variables {x -> y(x), y(x) -> x}
differential order: 1; trying a linearization to 2nd order
trying 1st order ODE linearizable_by_differentiation
--- Trying Lie symmetry methods, 1st order ---
`, `-> Computing symmetries using: way = 3`[x, x^2-1]
dsolve([diff(y(x),x)=x*(1-exp(2*y(x)-x^2)),y(0) = 0],y(x), singsol=all)
DSolve[{y'[x]==x*(1-Exp[2*y[x]-x^2]),{y[0]==0}},y[x],x,IncludeSingularSolutions -> True]