Internal problem ID [7439]
Internal file name [OUTPUT/6406_Sunday_June_05_2022_04_46_41_PM_5439210/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 50.
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]]
\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{3}=0} \]
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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{3} = 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^{2}}{y} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=p^{2}\). Integrating both sides gives \begin{align*} \frac {1}{p^{2}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{p^{2}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ -\frac {1}{p}&=-\ln \left (y \right )+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{p \left (y \right )}+\ln \left (y \right )-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {1}{y^{\prime }}+\ln \left (y\right )-c_{1} = 0 \end {align*}
Integrating both sides gives \begin {align*} \int \left (\ln \left (y \right )-c_{1} \right )d y &= x +c_{2}\\ \left (\ln \left (y \right )-c_{1} \right ) y -y&=x +c_{2} \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}^{\operatorname {LambertW}\left (\left (x +c_{2} \right ) {\mathrm e}^{-1-c_{1}}\right )+1+c_{1}} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\operatorname {LambertW}\left (\left (x +c_{2} \right ) {\mathrm e}^{-1-c_{1}}\right )+1+c_{1}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (\frac {d}{d x}y^{\prime }\right )+{y^{\prime }}^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\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 ),\frac {d}{d x}y^{\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 )+u \left (y \right )^{3}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {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 )^{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 )^{2}}d y =\int -\frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (y \right )}=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {1}{\ln \left (y \right )-c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {1}{\ln \left (y \right )-c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {1}{\ln \left (y\right )-c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{\ln \left (y\right )-c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\ln \left (y\right )-c_{1} \right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } \left (\ln \left (y\right )-c_{1} \right )d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -c_{1} y+y \ln \left (y\right )-y=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\mathit {LambertW}\left (\left (x +c_{2} \right ) {\mathrm e}^{-1-c_{1}}\right )+1+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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)^3/_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: 0.032 (sec). Leaf size: 27
dsolve(y(x)*diff(y(x),x$2)+diff(y(x),x)^3=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= \frac {x +c_{2}}{\operatorname {LambertW}\left (\left (x +c_{2} \right ) {\mathrm e}^{c_{1} -1}\right )} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.106 (sec). Leaf size: 26
DSolve[y[x]*y''[x]+y'[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x+c_2}{W\left (e^{-1-c_1} (x+c_2)\right )} \]