Internal problem ID [12494]
Internal file name [OUTPUT/11147_Monday_October_16_2023_09_52_00_PM_4354937/index.tex]
Book: DIFFERENTIAL and INTEGRAL CALCULUS. VOL I. by N. PISKUNOV. MIR
PUBLISHERS, Moscow 1969.
Section: Chapter 8. Differential equations. Exercises page 595
Problem number: 123.
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]]
\[ \boxed {y y^{\prime \prime }-{y^{\prime }}^{2}+{y^{\prime }}^{3}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 0] \end {align*}
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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-p \left (y \right )+p \left (y \right )^{2}\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 \right )+\ln \left (p -1\right )&=-\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (p \right )+\ln \left (p -1\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*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(y=-1\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = \frac {1}{c_{2} +1} \end {align*}
Unable to solve for constant of integration. Since \(\lim _{c_1 \to \infty }\) gives \(p = -\frac {y}{c_{2} -y}=p = 0\) and this result satisfies the given initial condition. 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 } = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1 = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = -1 \end {align*}
Trying the constant \begin {align*} c_{3} = -1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-1 \end {align*}
The constant \(c_{3} = -1\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -1 \\ \end{align*}
Verification of solutions
\[ y = -1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y y^{\prime \prime }+\left (-y^{\prime }+{y^{\prime }}^{2}\right ) y^{\prime }=0, y \left (0\right )=-1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \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} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+\left (-u \left (y \right )+u \left (y \right )^{2}\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 )+u \left (y \right )^{2}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{-u \left (y \right )+u \left (y \right )^{2}}=-\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 )+u \left (y \right )^{2}}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 (u \left (y \right )\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}}}} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} 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}}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1={\mathrm e}^{-\frac {\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2}}{{\mathrm e}^{c_{1}}}}\right ) {\mathrm e}^{c_{1}}-c_{2}}{{\mathrm e}^{c_{1}}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (-\frac {\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2} +x}{{\mathrm e}^{c_{1}}}}\right )}{1+\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2} +x}{{\mathrm e}^{c_{1}}}}\right )}+1\right ) {\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}}}}}{{\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {\left (-\frac {\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2}}{{\mathrm e}^{c_{1}}}}\right )}{1+\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2}}{{\mathrm e}^{c_{1}}}}\right )}+1\right ) {\mathrm e}^{-\frac {\mathit {LambertW}\left (-{\mathrm e}^{-\frac {c_{1} {\mathrm e}^{c_{1}}-c_{2}}{{\mathrm e}^{c_{1}}}}\right ) {\mathrm e}^{c_{1}}-c_{2}}{{\mathrm e}^{c_{1}}}}}{{\mathrm e}^{c_{1}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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: 0.063 (sec). Leaf size: 5
dsolve([y(x)*diff(y(x),x$2)-diff(y(x),x)^2+diff(y(x),x)^3=0,y(0) = -1, D(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = -1 \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y[x]*y''[x]-(y'[x])^2+(y'[x])^3==0,{y[0]==-1,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
{}