1.50 problem 50

Internal problem ID [8087]
Book : Second order enumerated odes
Section : section 1
Problem number : 50
Date solved : Monday, October 21, 2024 at 04:49:15 PM
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*}

1.50.1 Solved as second order missing x ode

Time used: 0.465 (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 \(p\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} p&=0\\ 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 \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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.

1.50.2 Maple step by step solution

\[ \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 +\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 )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {1}{\ln \left (y\right )-\mathit {C1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{\ln \left (y\right )-\mathit {C1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\ln \left (y\right )-\mathit {C1} \right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } \left (\ln \left (y\right )-\mathit {C1} \right )d x =\int 1d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -y \mathit {C1} +y \ln \left (y\right )-y=\mathit {C2} +x \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\mathit {LambertW}\left (\left (\mathit {C2} +x \right ) {\mathrm e}^{-\mathit {C1} -1}\right )+\mathit {C1} +1} \end {array} \]

1.50.3 Maple trace

Methods for second order ODEs:
 

1.50.4 Maple dsolve solution

Solving time : 0.017 (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 {c_2 +x}{\operatorname {LambertW}\left (\left (c_2 +x \right ) {\mathrm e}^{c_1 -1}\right )} \\ \end{align*}

1.50.5 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]
 

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