Internal problem ID [4662]
Internal file name [OUTPUT/4155_Sunday_June_05_2022_12_30_20_PM_45097950/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.12, page 504.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{3}-{y^{\prime }}^{2}=0} \]
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 )^{2}-p \left (y \right )\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 \left (p -1\right )}{y} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=p \left (p -1\right )\). Integrating both sides gives \begin{align*} \frac {1}{p \left (p -1\right )} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{p \left (p -1\right )} \,dp} &= \int {-\frac {1}{y} \,d y} \\ \ln \left (p -1\right )-\ln \left (p \right )&=-\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (p -1\right )-\ln \left (p \right )} &= {\mathrm e}^{-\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {p -1}{p} &= \frac {c_{2}}{y} \end {align*}
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 {y}{c_{2} -y} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {-c_{2} +y}{y}d y &= x +c_{3}\\ y -c_{2} \ln \left (y \right )&=x +c_{3} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\frac {c_{2} \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {x +c_{3}}{c_{2}}}}{c_{2}}\right )+c_{3} +x}{c_{2}}} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-\frac {c_{2} \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-\frac {x +c_{3}}{c_{2}}}}{c_{2}}\right )+c_{3} +x}{c_{2}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } y+\left ({y^{\prime }}^{2}-y^{\prime }\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 )^{2}-u \left (y \right )\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 )^{2}-u \left (y \right )}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )^{2}-u \left (y \right )}=-\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 )^{2}-u \left (y \right )}d y =\int -\frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (u \left (y \right )\right )+\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 } \left ({\mathrm e}^{c_{1}}-y\right )}{y}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } \left ({\mathrm e}^{c_{1}}-y\right )}{y}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -y+{\mathrm e}^{c_{1}} \ln \left (y\right )=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-\frac {\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2} +x}{{\mathrm e}^{c_{1}}}}\right ) {\mathrm e}^{c_{1}}-c_{2} +x}{{\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 a change of variables {x -> y(x), y(x) -> x} 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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)^2*(_b(_a)-1)/_a = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 36
dsolve(y(x)*diff(y(x),x$2)+(diff(y(x),x))^3-diff(y(x),x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= {\mathrm e}^{\frac {-c_{1} \operatorname {LambertW}\left (\frac {{\mathrm e}^{\frac {c_{2} +x}{c_{1}}}}{c_{1}}\right )+c_{2} +x}{c_{1}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 22.229 (sec). Leaf size: 32
DSolve[y[x]*y''[x]+(y'[x])^3-(y'[x])^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{c_1} W\left (e^{e^{-c_1} \left (x-e^{c_1} c_1+c_2\right )}\right ) \]