Internal problem ID [10021]
Internal file name [OUTPUT/8968_Monday_June_06_2022_06_05_27_AM_73994194/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1700 (book 6.109).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime } y+{y^{\prime }}^{2}-y^{\prime }=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime } y+\left (y^{\prime }-1\right ) y^{\prime }\right )d x &= 0 \\ y^{\prime } y-y = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {y}{y +c_{1}}d y &= x +c_{2}\\ y -c_{1} \ln \left (y +c_{1} \right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1} +c_{2} +x}{c_{1}}}}{c_{1}}\right )+1\right )\\ &=-c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{c_{1}}}}{c_{1} c_{2}}\right )+1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{c_{1}}}}{c_{1} c_{2}}\right )+1\right ) \\ \end{align*}
Verification of solutions
\[ y = -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{c_{1}}}}{c_{1} c_{2}}\right )+1\right ) \] Verified OK.
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y +\left (p \left (y \right )-1\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {-p +1}{y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=-p +1\). Integrating both sides gives \begin{align*} \frac {1}{-p +1} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{-p +1} \,dp} &= \int {\frac {1}{y} \,d y} \\ -\ln \left (p -1\right )&=\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{p -1} &= {\mathrm e}^{\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {1}{p -1} &= c_{2} y \end {align*}
Which simplifies to \[ p \left (y \right ) = \frac {\left (c_{2} y \,{\mathrm e}^{c_{1}}+1\right ) {\mathrm e}^{-c_{1}}}{c_{2} y} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = \frac {\left (c_{2} y \,{\mathrm e}^{c_{1}}+1\right ) {\mathrm e}^{-c_{1}}}{c_{2} y} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {c_{2} y \,{\mathrm e}^{c_{1}}}{c_{2} y \,{\mathrm e}^{c_{1}}+1}d y &= c_{3} +x\\ y -\frac {{\mathrm e}^{-c_{1}} \ln \left (c_{2} y \,{\mathrm e}^{c_{1}}+1\right )}{c_{2}}&=c_{3} +x \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {\left (\operatorname {LambertW}\left (-{\mathrm e}^{-c_{3} c_{2} {\mathrm e}^{c_{1}}-x c_{2} {\mathrm e}^{c_{1}}-1}\right )+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}}\\ &=-\frac {\left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-x c_{2} {\mathrm e}^{c_{1}}} {\mathrm e}^{-1}}{c_{3}}\right )+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-x c_{2} {\mathrm e}^{c_{1}}} {\mathrm e}^{-1}}{c_{3}}\right )+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-x c_{2} {\mathrm e}^{c_{1}}} {\mathrm e}^{-1}}{c_{3}}\right )+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \] Verified OK.
Writing the ode as \[ y^{\prime \prime } y+\left (y^{\prime }-1\right ) y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime } y+\left (y^{\prime }-1\right ) y^{\prime }\right )d x &= 0 \\ y^{\prime } y-y = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {y}{y +c_{1}}d y &= x +c_{2}\\ y -c_{1} \ln \left (y +c_{1} \right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {c_{1} +c_{2} +x}{c_{1}}}}{c_{1}}\right )+1\right )\\ &=-c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{c_{1}}}}{c_{1} c_{2}}\right )+1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{c_{1}}}}{c_{1} c_{2}}\right )+1\right ) \\ \end{align*}
Verification of solutions
\[ y = -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} {\mathrm e}^{-\frac {x}{c_{1}}}}{c_{1} c_{2}}\right )+1\right ) \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= y\\ a_1 &= y^{\prime }-1\\ a_0 &= 0 \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {y\,d y'} + \int {y^{\prime }-1\,d y} + \int {0\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} y^{\prime } y+y \left (y^{\prime }-1\right ) = c_{1} \end {align*}
Which is now solved Integrating both sides gives \begin {align*} \int \frac {2 y}{y +c_{1}}d y &= x +c_{2}\\ 2 y -2 c_{1} \ln \left (y +c_{1} \right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {2 c_{1} +c_{2} +x}{2 c_{1}}}}{c_{1}}\right )+1\right )\\ &=-c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} c_{2} {\mathrm e}^{-\frac {x}{2 c_{1}}}}{c_{1}}\right )+1\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} c_{2} {\mathrm e}^{-\frac {x}{2 c_{1}}}}{c_{1}}\right )+1\right ) \\ \end{align*}
Verification of solutions
\[ y = -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-1} c_{2} {\mathrm e}^{-\frac {x}{2 c_{1}}}}{c_{1}}\right )+1\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } y+\left (y^{\prime }-1\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right ) y +\left (u \left (y \right )-1\right ) u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {u \left (y \right )-1}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )-1}=-\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )-1}d y =\int -\frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )-1\right )=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {y +{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {y +{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {y+{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y+{\mathrm e}^{c_{1}}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{y+{\mathrm e}^{c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{y+{\mathrm e}^{c_{1}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y-{\mathrm e}^{c_{1}} \ln \left (y+{\mathrm e}^{c_{1}}\right )=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}+{\mathrm e}^{c_{1}}+c_{2} +x}{{\mathrm e}^{c_{1}}}}\right ) {\mathrm e}^{c_{1}}-{\mathrm e}^{c_{1}} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)*(_b(_a)-1)/_a = 0, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 34
dsolve(diff(diff(y(x),x),x)*y(x)+diff(y(x),x)^2-diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= -c_{1} \left (\operatorname {LambertW}\left (-\frac {{\mathrm e}^{\frac {-c_{1} -c_{2} -x}{c_{1}}}}{c_{1}}\right )+1\right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.142 (sec). Leaf size: 32
DSolve[-y'[x] + y'[x]^2 + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -c_1 \left (1+W\left (-\frac {e^{-\frac {x+c_1+c_2}{c_1}}}{c_1}\right )\right ) \]