1.46 problem 46

Internal problem ID [8083]
Book : Second order enumerated odes
Section : section 1
Problem number : 46
Date solved : Monday, October 21, 2024 at 04:47:58 PM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y {y^{\prime \prime }}^{2}+{y^{\prime }}^{3}&=0 \end{align*}

1.46.1 Solved as second order missing x ode

Time used: 2.108 (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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right )^{3} = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

Solving for \(p^{\prime }\) gives

\begin{align*} p^{\prime }&=\frac {\sqrt {-p y}}{y}\tag {1} \\ p^{\prime }&=-\frac {\sqrt {-p y}}{y}\tag {2} \end{align*}

In canonical form, the ODE is

\begin{align*} p' &= F(y,p)\\ &= \frac {\sqrt {-p y}}{y}\tag {1} \end{align*}

An ode of the form \(p' = \frac {M(y,p)}{N(y,p)}\) is called homogeneous if the functions \(M(y,p)\) and \(N(y,p)\) are both homogeneous functions and of the same order. Recall that a function \(f(y,p)\) is homogeneous of order \(n\) if

\[ f(t^n y, t^n p)= t^n f(y,p) \]

In this case, it can be seen that both \(M=\sqrt {-p y}\) and \(N=y\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {p}{y}\), or \(p=uy\). Hence

\[ \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}y}}y + u \]

Applying the transformation \(p=uy\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}y}}y + u &= \sqrt {-u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}y}} &= \frac {\sqrt {-u \left (y \right )}-u \left (y \right )}{y} \end{align*}

Or

\[ u^{\prime }\left (y \right )-\frac {\sqrt {-u \left (y \right )}-u \left (y \right )}{y} = 0 \]

Or

\[ u^{\prime }\left (y \right ) y -\sqrt {-u \left (y \right )}+u \left (y \right ) = 0 \]

Which is now solved as separable in \(u \left (y \right )\).

The ode \(u^{\prime }\left (y \right ) = \frac {\sqrt {-u \left (y \right )}-u \left (y \right )}{y}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (y \right )&= \frac {\sqrt {-u \left (y \right )}-u \left (y \right )}{y}\\ &= f(y) g(u) \end{align*}

Where

\begin{align*} f(y) &= \frac {1}{y}\\ g(u) &= \sqrt {-u}-u \end{align*}

Integrating gives

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

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

\begin{align*} u \left (y \right )&=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*} \ln \left (\frac {1}{\left (\sqrt {-u \left (y \right )}+1\right )^{2}}\right ) = \ln \left (y \right )+c_1\\ u \left (y \right ) = 0 \end{align*}

Converting \(\ln \left (\frac {1}{\left (\sqrt {-u \left (y \right )}+1\right )^{2}}\right ) = \ln \left (y \right )+c_1\) back to \(p\) gives

\begin{align*} \ln \left (\frac {1}{\left (\sqrt {-\frac {p}{y}}+1\right )^{2}}\right ) = \ln \left (y \right )+c_1 \end{align*}

Converting \(u \left (y \right ) = 0\) back to \(p\) gives

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

In canonical form, the ODE is

\begin{align*} p' &= F(y,p)\\ &= -\frac {\sqrt {-p y}}{y}\tag {1} \end{align*}

An ode of the form \(p' = \frac {M(y,p)}{N(y,p)}\) is called homogeneous if the functions \(M(y,p)\) and \(N(y,p)\) are both homogeneous functions and of the same order. Recall that a function \(f(y,p)\) is homogeneous of order \(n\) if

\[ f(t^n y, t^n p)= t^n f(y,p) \]

In this case, it can be seen that both \(M=-\sqrt {-p y}\) and \(N=y\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {p}{y}\), or \(p=uy\). Hence

\[ \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}y}}y + u \]

Applying the transformation \(p=uy\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}y}}y + u &= -\sqrt {-u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}y}} &= \frac {-\sqrt {-u \left (y \right )}-u \left (y \right )}{y} \end{align*}

Or

\[ u^{\prime }\left (y \right )-\frac {-\sqrt {-u \left (y \right )}-u \left (y \right )}{y} = 0 \]

Or

\[ u^{\prime }\left (y \right ) y +\sqrt {-u \left (y \right )}+u \left (y \right ) = 0 \]

Which is now solved as separable in \(u \left (y \right )\).

The ode \(u^{\prime }\left (y \right ) = -\frac {\sqrt {-u \left (y \right )}+u \left (y \right )}{y}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (y \right )&= -\frac {\sqrt {-u \left (y \right )}+u \left (y \right )}{y}\\ &= f(y) g(u) \end{align*}

Where

\begin{align*} f(y) &= \frac {1}{y}\\ g(u) &= -\sqrt {-u}-u \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(y) \,dy}\\ \int { \frac {1}{-\sqrt {-u}-u}\,du} &= \int { \frac {1}{y} \,dy}\\ -2 \ln \left (\sqrt {-u \left (y \right )}-1\right )&=\ln \left (y \right )+c_2 \end{align*}

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

\begin{align*} u \left (y \right )&=-1\\ u \left (y \right )&=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*} -2 \ln \left (\sqrt {-u \left (y \right )}-1\right ) = \ln \left (y \right )+c_2\\ u \left (y \right ) = -1\\ u \left (y \right ) = 0 \end{align*}

Converting \(-2 \ln \left (\sqrt {-u \left (y \right )}-1\right ) = \ln \left (y \right )+c_2\) back to \(p\) gives

\begin{align*} -2 \ln \left (\sqrt {-\frac {p}{y}}-1\right ) = \ln \left (y \right )+c_2 \end{align*}

Converting \(u \left (y \right ) = -1\) back to \(p\) gives

\begin{align*} p = -y \end{align*}

Converting \(u \left (y \right ) = 0\) back to \(p\) gives

\begin{align*} 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&=\left (-\frac {2 y \,{\mathrm e}^{c_1} \left (\sqrt {y \,{\mathrm e}^{c_1}}-1\right )}{\sqrt {y \,{\mathrm e}^{c_1}}}+y \,{\mathrm e}^{c_1}-1\right ) {\mathrm e}^{-c_1}\\ p&=\left (-\frac {2 y \,{\mathrm e}^{c_1} \left (\sqrt {y \,{\mathrm e}^{c_1}}+1\right )}{\sqrt {y \,{\mathrm e}^{c_1}}}+y \,{\mathrm e}^{c_1}-1\right ) {\mathrm e}^{-c_1}\\ p&=-y\\ p&=-\left (\frac {2 y \,{\mathrm e}^{c_2} \left (\sqrt {y \,{\mathrm e}^{c_2}}-1\right )}{\sqrt {y \,{\mathrm e}^{c_2}}}-y \,{\mathrm e}^{c_2}+1\right ) {\mathrm e}^{-c_2}\\ p&=-\left (\frac {2 y \,{\mathrm e}^{c_2} \left (\sqrt {y \,{\mathrm e}^{c_2}}+1\right )}{\sqrt {y \,{\mathrm e}^{c_2}}}-y \,{\mathrm e}^{c_2}+1\right ) {\mathrm e}^{-c_2} \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_3 \\ y &= c_3 \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 } = \left (-\frac {2 y \,{\mathrm e}^{c_1} \left (\sqrt {y \,{\mathrm e}^{c_1}}-1\right )}{\sqrt {y \,{\mathrm e}^{c_1}}}+y \,{\mathrm e}^{c_1}-1\right ) {\mathrm e}^{-c_1} \end{align*}

Integrating gives

\begin{align*} \int -\frac {\sqrt {y \,{\mathrm e}^{c_1}}\, {\mathrm e}^{c_1}}{y \,{\mathrm e}^{c_1} \sqrt {y \,{\mathrm e}^{c_1}}-2 y \,{\mathrm e}^{c_1}+\sqrt {y \,{\mathrm e}^{c_1}}}d y &= dx\\ \frac {2+\left (-2 \sqrt {y \,{\mathrm e}^{c_1}}+2\right ) \ln \left (\sqrt {y \,{\mathrm e}^{c_1}}-1\right )}{\sqrt {y \,{\mathrm e}^{c_1}}-1}&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {\left (y \,{\mathrm e}^{c_1} \sqrt {y \,{\mathrm e}^{c_1}}-2 y \,{\mathrm e}^{c_1}+\sqrt {y \,{\mathrm e}^{c_1}}\right ) {\mathrm e}^{-c_1}}{\sqrt {y \,{\mathrm e}^{c_1}}}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{-c_1} \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&={\mathrm e}^{-c_1}\\ y&=\frac {\left (\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )^{2}+2 \operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )+1\right ) {\mathrm e}^{-c_1}}{\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_4}{2}+\frac {x}{2}}\right )^{2}} \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \left (-\frac {2 y \,{\mathrm e}^{c_1} \left (1+\sqrt {y \,{\mathrm e}^{c_1}}\right )}{\sqrt {y \,{\mathrm e}^{c_1}}}+y \,{\mathrm e}^{c_1}-1\right ) {\mathrm e}^{-c_1} \end{align*}

Integrating gives

\begin{align*} \int -\frac {\sqrt {y \,{\mathrm e}^{c_1}}\, {\mathrm e}^{c_1}}{y \,{\mathrm e}^{c_1} \sqrt {y \,{\mathrm e}^{c_1}}+2 y \,{\mathrm e}^{c_1}+\sqrt {y \,{\mathrm e}^{c_1}}}d y &= dx\\ \frac {-2+\left (-2 \sqrt {y \,{\mathrm e}^{c_1}}-2\right ) \ln \left (\sqrt {y \,{\mathrm e}^{c_1}}+1\right )}{\sqrt {y \,{\mathrm e}^{c_1}}+1}&= x +c_5 \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y = \frac {\left (\operatorname {LambertW}\left (-{\mathrm e}^{\frac {c_5}{2}+\frac {x}{2}}\right )^{2}+2 \operatorname {LambertW}\left (-{\mathrm e}^{\frac {c_5}{2}+\frac {x}{2}}\right )+1\right ) {\mathrm e}^{-c_1}}{\operatorname {LambertW}\left (-{\mathrm e}^{\frac {c_5}{2}+\frac {x}{2}}\right )^{2}} \end{align*}

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -y \end{align*}

Integrating gives

\begin{align*} \int -\frac {1}{y}d y &= dx\\ -\ln \left (y \right )&= x +c_6 \end{align*}

Singular solutions are found by solving

\begin{align*} -y&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = 0 \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=0\\ y&={\mathrm e}^{-x -c_6} \end{align*}

For solution (5) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\left (\frac {2 y \,{\mathrm e}^{c_2} \left (\sqrt {y \,{\mathrm e}^{c_2}}-1\right )}{\sqrt {y \,{\mathrm e}^{c_2}}}-y \,{\mathrm e}^{c_2}+1\right ) {\mathrm e}^{-c_2} \end{align*}

Integrating gives

\begin{align*} \int -\frac {\sqrt {y \,{\mathrm e}^{c_2}}\, {\mathrm e}^{c_2}}{y \,{\mathrm e}^{c_2} \sqrt {y \,{\mathrm e}^{c_2}}-2 y \,{\mathrm e}^{c_2}+\sqrt {y \,{\mathrm e}^{c_2}}}d y &= dx\\ \frac {2+\left (-2 \sqrt {y \,{\mathrm e}^{c_2}}+2\right ) \ln \left (\sqrt {y \,{\mathrm e}^{c_2}}-1\right )}{\sqrt {y \,{\mathrm e}^{c_2}}-1}&= x +c_7 \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {\left (y \,{\mathrm e}^{c_2} \sqrt {y \,{\mathrm e}^{c_2}}-2 y \,{\mathrm e}^{c_2}+\sqrt {y \,{\mathrm e}^{c_2}}\right ) {\mathrm e}^{-c_2}}{\sqrt {y \,{\mathrm e}^{c_2}}}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{-c_2} \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&={\mathrm e}^{-c_2}\\ y&=\frac {\left (\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_7}{2}+\frac {x}{2}}\right )^{2}+2 \operatorname {LambertW}\left ({\mathrm e}^{\frac {c_7}{2}+\frac {x}{2}}\right )+1\right ) {\mathrm e}^{-c_2}}{\operatorname {LambertW}\left ({\mathrm e}^{\frac {c_7}{2}+\frac {x}{2}}\right )^{2}} \end{align*}

For solution (6) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\left (\frac {2 y \,{\mathrm e}^{c_2} \left (1+\sqrt {y \,{\mathrm e}^{c_2}}\right )}{\sqrt {y \,{\mathrm e}^{c_2}}}-y \,{\mathrm e}^{c_2}+1\right ) {\mathrm e}^{-c_2} \end{align*}

Integrating gives

\begin{align*} \int -\frac {\sqrt {y \,{\mathrm e}^{c_2}}\, {\mathrm e}^{c_2}}{y \,{\mathrm e}^{c_2} \sqrt {y \,{\mathrm e}^{c_2}}+2 y \,{\mathrm e}^{c_2}+\sqrt {y \,{\mathrm e}^{c_2}}}d y &= dx\\ \frac {-2+\left (-2 \sqrt {y \,{\mathrm e}^{c_2}}-2\right ) \ln \left (\sqrt {y \,{\mathrm e}^{c_2}}+1\right )}{\sqrt {y \,{\mathrm e}^{c_2}}+1}&= x +c_8 \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y = \frac {\left (\operatorname {LambertW}\left (-{\mathrm e}^{\frac {c_8}{2}+\frac {x}{2}}\right )^{2}+2 \operatorname {LambertW}\left (-{\mathrm e}^{\frac {c_8}{2}+\frac {x}{2}}\right )+1\right ) {\mathrm e}^{-c_2}}{\operatorname {LambertW}\left (-{\mathrm e}^{\frac {c_8}{2}+\frac {x}{2}}\right )^{2}} \end{align*}

Will add steps showing solving for IC soon.

1.46.2 Maple step by step solution

1.46.3 Maple trace

Methods for second order ODEs:
 

1.46.4 Maple dsolve solution

Solving time : 0.111 (sec)
Leaf size : 178

dsolve(y(x)*diff(diff(y(x),x),x)^2+diff(y(x),x)^3 = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= c_1 \\ y &= 0 \\ y &= {\mathrm e}^{-\left (\int {\mathrm e}^{2 \operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )+c_1 \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}-\ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )-c_1 +2 \textit {\_Z} -x +2\right )}d x \right )+2 \left (\int {\mathrm e}^{\operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )+c_1 \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}-\ln \left (\left ({\mathrm e}^{\textit {\_Z}}-1\right )^{2}\right )-c_1 +2 \textit {\_Z} -x +2\right )}d x \right )-x +c_2} \\ y &= \frac {c_2 {\left (\operatorname {LambertW}\left (c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )+1\right )}^{2}}{\operatorname {LambertW}\left (c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )^{2}} \\ y &= \frac {c_2 {\left (\operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )+1\right )}^{2}}{\operatorname {LambertW}\left (-c_1 \,{\mathrm e}^{-1+\frac {x}{2}}\right )^{2}} \\ \end{align*}

1.46.5 Mathematica DSolve solution

Solving time : 3.059 (sec)
Leaf size : 361

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

\begin{align*} y(x)\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i c_1\right )-\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}-i c_1\right )}\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}+i c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i c_1\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i (-c_1)\right )-\frac {i (-c_1)}{2 \left (2 \sqrt {\text {$\#$1}}-i (-c_1)\right )}\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {i (-c_1)}{2 \left (2 \sqrt {\text {$\#$1}}+i (-1) c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i (-1) c_1\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i c_1\right )-\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}-i c_1\right )}\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}+i c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i c_1\right )\right )\&\right ][x+c_2] \\ \end{align*}