Internal problem ID [8547]
Internal file name [OUTPUT/7480_Sunday_June_05_2022_10_57_38_PM_31217802/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 211.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "dAlembert", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _dAlembert]
\[ \boxed {y y^{\prime }-x \,{\mathrm e}^{\frac {x}{y}}=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} u \left (x \right ) x \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )-x \,{\mathrm e}^{\frac {1}{u \left (x \right )}} = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {-u^{2}+{\mathrm e}^{\frac {1}{u}}}{u x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {-u^{2}+{\mathrm e}^{\frac {1}{u}}}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {-u^{2}+{\mathrm e}^{\frac {1}{u}}}{u}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {-u^{2}+{\mathrm e}^{\frac {1}{u}}}{u}} \,du} &= \int {\frac {1}{x} \,d x} \\ \int _{}^{u}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a}&=\ln \left (x \right )+c_{2} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a}=\ln \left (x \right )+c_{2} \] The solution is \[ \int _{}^{u \left (x \right )}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \int _{}^{\frac {y}{x}}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0\\ \int _{}^{\frac {y}{x}}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{\frac {y}{x}}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a} -\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \int _{}^{\frac {y}{x}}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a} -\ln \left (x \right )-c_{2} = 0 \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=\frac {x \,{\mathrm e}^{\frac {x}{y}}}{y}\\ 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 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 {x \,{\mathrm e}^{\frac {x}{y}} \left (b_{3}-a_{2}\right )}{y}-\frac {x^{2} {\mathrm e}^{\frac {2 x}{y}} a_{3}}{y^{2}}-\left (\frac {{\mathrm e}^{\frac {x}{y}}}{y}+\frac {x \,{\mathrm e}^{\frac {x}{y}}}{y^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {x^{2} {\mathrm e}^{\frac {x}{y}}}{y^{3}}-\frac {x \,{\mathrm e}^{\frac {x}{y}}}{y^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {x^{2} {\mathrm e}^{\frac {2 x}{y}} a_{3} y -{\mathrm e}^{\frac {x}{y}} x^{3} b_{2}+{\mathrm e}^{\frac {x}{y}} x^{2} y a_{2}-{\mathrm e}^{\frac {x}{y}} x^{2} y b_{2}-{\mathrm e}^{\frac {x}{y}} x^{2} y b_{3}+2 \,{\mathrm e}^{\frac {x}{y}} x \,y^{2} a_{2}+{\mathrm e}^{\frac {x}{y}} x \,y^{2} a_{3}-2 \,{\mathrm e}^{\frac {x}{y}} x \,y^{2} b_{3}+{\mathrm e}^{\frac {x}{y}} y^{3} a_{3}-{\mathrm e}^{\frac {x}{y}} x^{2} b_{1}+{\mathrm e}^{\frac {x}{y}} x y a_{1}-{\mathrm e}^{\frac {x}{y}} x y b_{1}+{\mathrm e}^{\frac {x}{y}} y^{2} a_{1}-b_{2} y^{3}}{y^{3}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -x^{2} {\mathrm e}^{\frac {2 x}{y}} a_{3} y +{\mathrm e}^{\frac {x}{y}} x^{3} b_{2}-{\mathrm e}^{\frac {x}{y}} x^{2} y a_{2}+{\mathrm e}^{\frac {x}{y}} x^{2} y b_{2}+{\mathrm e}^{\frac {x}{y}} x^{2} y b_{3}-2 \,{\mathrm e}^{\frac {x}{y}} x \,y^{2} a_{2}-{\mathrm e}^{\frac {x}{y}} x \,y^{2} a_{3}+2 \,{\mathrm e}^{\frac {x}{y}} x \,y^{2} b_{3}-{\mathrm e}^{\frac {x}{y}} y^{3} a_{3}+{\mathrm e}^{\frac {x}{y}} x^{2} b_{1}-{\mathrm e}^{\frac {x}{y}} x y a_{1}+{\mathrm e}^{\frac {x}{y}} x y b_{1}-{\mathrm e}^{\frac {x}{y}} y^{2} a_{1}+b_{2} y^{3} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} -x^{2} {\mathrm e}^{\frac {2 x}{y}} a_{3} y +{\mathrm e}^{\frac {x}{y}} x^{3} b_{2}-{\mathrm e}^{\frac {x}{y}} x^{2} y a_{2}+{\mathrm e}^{\frac {x}{y}} x^{2} y b_{2}+{\mathrm e}^{\frac {x}{y}} x^{2} y b_{3}-2 \,{\mathrm e}^{\frac {x}{y}} x \,y^{2} a_{2}-{\mathrm e}^{\frac {x}{y}} x \,y^{2} a_{3}+2 \,{\mathrm e}^{\frac {x}{y}} x \,y^{2} b_{3}-{\mathrm e}^{\frac {x}{y}} y^{3} a_{3}+{\mathrm e}^{\frac {x}{y}} x^{2} b_{1}-{\mathrm e}^{\frac {x}{y}} x y a_{1}+{\mathrm e}^{\frac {x}{y}} x y b_{1}-{\mathrm e}^{\frac {x}{y}} y^{2} a_{1}+b_{2} y^{3} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, {\mathrm e}^{\frac {x}{y}}, {\mathrm e}^{\frac {2 x}{y}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\mathrm e}^{\frac {x}{y}} = v_{3}, {\mathrm e}^{\frac {2 x}{y}} = v_{4}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -v_{3} v_{1}^{2} v_{2} a_{2}-2 v_{3} v_{1} v_{2}^{2} a_{2}-v_{1}^{2} v_{4} a_{3} v_{2}-v_{3} v_{1} v_{2}^{2} a_{3}-v_{3} v_{2}^{3} a_{3}+v_{3} v_{1}^{3} b_{2}+v_{3} v_{1}^{2} v_{2} b_{2}+v_{3} v_{1}^{2} v_{2} b_{3}+2 v_{3} v_{1} v_{2}^{2} b_{3}-v_{3} v_{1} v_{2} a_{1}-v_{3} v_{2}^{2} a_{1}+v_{3} v_{1}^{2} b_{1}+v_{3} v_{1} v_{2} b_{1}+b_{2} v_{2}^{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} v_{3} v_{1}^{3} b_{2}+\left (-a_{2}+b_{2}+b_{3}\right ) v_{1}^{2} v_{2} v_{3}-v_{1}^{2} v_{4} a_{3} v_{2}+v_{3} v_{1}^{2} b_{1}+\left (-2 a_{2}-a_{3}+2 b_{3}\right ) v_{1} v_{2}^{2} v_{3}+\left (-a_{1}+b_{1}\right ) v_{1} v_{2} v_{3}-v_{3} v_{2}^{3} a_{3}+b_{2} v_{2}^{3}-v_{3} v_{2}^{2} a_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} b_{1}&=0\\ b_{2}&=0\\ -a_{1}&=0\\ -a_{3}&=0\\ -a_{1}+b_{1}&=0\\ -2 a_{2}-a_{3}+2 b_{3}&=0\\ -a_{2}+b_{2}+b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \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 &= y \\ \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 \(S(R)\). Unable to determine \(R\). Terminating
Let \(p=y^{\prime }\) the ode becomes \begin {align*} y p -x \,{\mathrm e}^{\frac {x}{y}} = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \frac {x}{\operatorname {LambertW}\left (p \right )}\tag {1A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved. Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= \frac {1}{\operatorname {LambertW}\left (p \right )}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {1}{\operatorname {LambertW}\left (p \right )} = -\frac {x p^{\prime }\left (x \right )}{\operatorname {LambertW}\left (p \right ) \left (1+\operatorname {LambertW}\left (p \right )\right ) p}\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -\frac {1}{\operatorname {LambertW}\left (p \right )} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=\frac {1}{2 \operatorname {LambertW}\left (\frac {1}{2}\right )}\\ p&=\frac {1}{2 \operatorname {LambertW}\left (-\frac {1}{2}\right )} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=\frac {x}{\operatorname {LambertW}\left (\frac {1}{2 \operatorname {LambertW}\left (-\frac {1}{2}\right )}\right )}\\ y&=\frac {x}{\operatorname {LambertW}\left (\frac {1}{2 \operatorname {LambertW}\left (\frac {1}{2}\right )}\right )} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = -\frac {\left (p \left (x \right )-\frac {1}{\operatorname {LambertW}\left (p \left (x \right )\right )}\right ) \operatorname {LambertW}\left (p \left (x \right )\right ) \left (1+\operatorname {LambertW}\left (p \left (x \right )\right )\right ) p \left (x \right )}{x}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = -\frac {x \left (p \right )}{\operatorname {LambertW}\left (p \right ) \left (1+\operatorname {LambertW}\left (p \right )\right ) p \left (p -\frac {1}{\operatorname {LambertW}\left (p \right )}\right )}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=\frac {1}{p \left (1+\operatorname {LambertW}\left (p \right )\right ) \left (p \operatorname {LambertW}\left (p \right )-1\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )+\frac {x \left (p \right )}{\left (1+\operatorname {LambertW}\left (p \right )\right ) p \left (p \operatorname {LambertW}\left (p \right )-1\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {1}{p \left (1+\operatorname {LambertW}\left (p \right )\right ) \left (p \operatorname {LambertW}\left (p \right )-1\right )}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int \frac {1}{p \left (1+\operatorname {LambertW}\left (p \right )\right ) \left (p \operatorname {LambertW}\left (p \right )-1\right )}d p} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int \frac {1}{p \left (1+\operatorname {LambertW}\left (p \right )\right ) \left (p \operatorname {LambertW}\left (p \right )-1\right )}d p} x &= c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {1}{p \left (1+\operatorname {LambertW}\left (p \right )\right ) \left (p \operatorname {LambertW}\left (p \right )-1\right )}d p}\) results in \begin {align*} x \left (p \right ) &= c_{2} {\mathrm e}^{-\left (\int \frac {1}{p \left (1+\operatorname {LambertW}\left (p \right )\right ) \left (p \operatorname {LambertW}\left (p \right )-1\right )}d p \right )} \end {align*}
Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x}{\operatorname {LambertW}\left (\frac {1}{2 \operatorname {LambertW}\left (-\frac {1}{2}\right )}\right )} \\ \tag{2} y &= \frac {x}{\operatorname {LambertW}\left (\frac {1}{2 \operatorname {LambertW}\left (\frac {1}{2}\right )}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x}{\operatorname {LambertW}\left (\frac {1}{2 \operatorname {LambertW}\left (-\frac {1}{2}\right )}\right )} \] Verified OK.
\[ y = \frac {x}{\operatorname {LambertW}\left (\frac {1}{2 \operatorname {LambertW}\left (\frac {1}{2}\right )}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y y^{\prime }-x \,{\mathrm e}^{\frac {x}{y}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x \,{\mathrm e}^{\frac {x}{y}}}{y} \end {array} \]
Maple trace
`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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 31
dsolve(y(x)*diff(y(x),x)-x*exp(x/y(x))=0,y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {\textit {\_a}}{-\textit {\_a}^{2}+{\mathrm e}^{\frac {1}{\textit {\_a}}}}d \textit {\_a} \right )+\ln \left (x \right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.218 (sec). Leaf size: 41
DSolve[y[x]*y'[x]-x*Exp[x/y[x]]==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{\frac {y(x)}{x}}\frac {K[1]}{K[1]^2-e^{\frac {1}{K[1]}}}dK[1]=-\log (x)+c_1,y(x)\right ] \]