2.1.50 problem 50

Internal problem ID [8797]
Book : Second order enumerated odes
Section : section 1
Problem number : 50
Date solved : Thursday, December 12, 2024 at 09:49:06 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_y_y1]]

Solve

\begin{align*} y y^{\prime \prime }+{y^{\prime }}^{3}&=0 \end{align*}

Solved as second order missing x ode

Time used: 0.480 (sec)

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 \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= 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 \(p(y)\).

The ode \(p^{\prime } = -\frac {p^{2}}{y}\) is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {p^{2}}{y}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -\frac {1}{y}\\ g(p) &= p^{2} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { \frac {1}{p^{2}}\,dp} &= \int { -\frac {1}{y} \,dy}\\ -\frac {1}{p}&=\ln \left (\frac {1}{y}\right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or \(p^{2}=0\) for \(p\) gives

\begin{align*} p&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {1}{p} = \ln \left (\frac {1}{y}\right )+c_1\\ p = 0 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= -\frac {1}{\ln \left (\frac {1}{y}\right )+c_1} \\ \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 } = 0 \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + c_2 \\ y &= c_2 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\frac {1}{\ln \left (\frac {1}{y}\right )+c_1} \end{align*}

Integrating gives

\begin{align*} \int \left (-\ln \left (\frac {1}{y}\right )-c_1 \right )d y &= dx\\ -y \left (\ln \left (\frac {1}{y}\right )+c_1 +1\right )&= x +c_3 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {x +c_3}{\operatorname {LambertW}\left (\left (x +c_3 \right ) {\mathrm e}^{-c_1 -1}\right )} \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_2 \\ y &= \frac {x +c_3}{\operatorname {LambertW}\left (\left (x +c_3 \right ) {\mathrm e}^{-c_1 -1}\right )} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) y \left (x \right )+\left (\frac {d}{d x}y \left (x \right )\right )^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} \frac {d}{d x}y \left (x \right )=u \left (y \right ),\frac {d^{2}}{d x^{2}}y \left (x \right )=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 +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 +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (y \right )}=-\ln \left (y \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {1}{\ln \left (y \right )-\mathit {C1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {1}{\ln \left (y \right )-\mathit {C1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {1}{\ln \left (y \left (x \right )\right )-\mathit {C1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {1}{\ln \left (y \left (x \right )\right )-\mathit {C1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) \left (\ln \left (y \left (x \right )\right )-\mathit {C1} \right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right ) \left (\ln \left (y \left (x \right )\right )-\mathit {C1} \right )d x =\int 1d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\mathit {C1} y \left (x \right )+y \left (x \right ) \ln \left (y \left (x \right )\right )-y \left (x \right )=x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{\mathit {LambertW}\left (\left (x +\mathit {C2} \right ) {\mathrm e}^{-\mathit {C1} -1}\right )+\mathit {C1} +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 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`
 

Maple dsolve solution

Solving time : 0.037 (sec)
Leaf size : 27

dsolve(y(x)*diff(diff(y(x),x),x)+diff(y(x),x)^3 = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= 0 \\ y &= c_{1} \\ y &= \frac {x +c_{2}}{\operatorname {LambertW}\left (\left (x +c_{2} \right ) {\mathrm e}^{c_{1} -1}\right )} \\ \end{align*}

Mathematica DSolve solution

Solving time : 60.147 (sec)
Leaf size : 26

DSolve[{y[x]*D[y[x],{x,2}]+D[y[x],x]^3==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \frac {x+c_2}{W\left (e^{-1-c_1} (x+c_2)\right )} \]