2.1.46 Problem 46

Internal problem ID [9117]
Book : Second order enumerated odes
Section : section 1
Problem number : 46
Date solved : Sunday, March 30, 2025 at 02:07:39 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order missing x ode

Time used: 2.856 (sec)

Solve

$$\begin{array}{rl}y{y^{\prime \prime }}^2+{y^{\prime }}^3& =0\end{array}$$

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

$$\begin{array}{rl}{y}^{\prime }& =p\end{array}$$

Then

$$\begin{array}{rl}{y}^{″}& =\frac{dp}{dx}\\ & =\frac{dp}{dy}\frac{dy}{dx}\\ & =p\frac{dp}{dy}\end{array}$$

Hence the ode becomes

$$\begin{array}{r}yp{\left(y\right)}^2{\left(\frac{d}{dy}p\left(y\right)\right)}^2+p{\left(y\right)}^3=0\end{array}$$

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

Solving for $p^{\prime }$ gives

$$\begin{array}{}\text{(1)}& p^{\prime }& =\frac{\sqrt{-py}}{y}\text{(2)}& p^{\prime }& =-\frac{\sqrt{-py}}{y}\end{array}$$

Solving ODE (1)

In canonical form, the ODE is

$$\begin{array}{rl}{p}^{\prime }& =F(y,p)\\ \text{(1)}& & =\frac{\sqrt{-py}}{y}\end{array}$$

An ode of the form $p^{\prime }=\frac{M(y,p)}{N(y,p)}$ 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 $M=\sqrt{-py}$ 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

$$\frac{\mathrm{d}p}{\mathrm{d}y}=\frac{\mathrm{d}u}{\mathrm{d}y}y+u$$

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

$$\begin{array}{rl}\frac{\mathrm{d}u}{\mathrm{d}y}y+u& =\sqrt{-u}\\ \frac{\mathrm{d}u}{\mathrm{d}y}& =\frac{\sqrt{-u\left(y\right)}-u\left(y\right)}{y}\end{array}$$

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

$$\begin{array}{}\text{(1)}& u^{\prime }\left(y\right)=\frac{\sqrt{-u\left(y\right)}-u\left(y\right)}{y}\end{array}$$

is separable as it can be written as

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

Where

$$\begin{array}{rl}f(y)& =\frac{1}{y}\\ g(u)& =\sqrt{-u}-u\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(y)dy\\ \int \frac{1}{\sqrt{-u}-u}du& =\int \frac{1}{y}dy\end{array}$$

$$\ln \left(\frac{1}{{(\sqrt{-u\left(y\right)}+1)}^2}\right)=\ln \left(y\right)+c_1$$

Taking the exponential of both sides the solution becomes

$$\frac{1}{{(\sqrt{-u\left(y\right)}+1)}^2}=c_{1}y$$

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

$$\begin{array}{rl}u\left(y\right)& =0\end{array}$$

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

$$\begin{array}{rl}\frac{1}{{(\sqrt{-u\left(y\right)}+1)}^2}& =c_{1}y\\ u\left(y\right)& =0\end{array}$$

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

$$\begin{array}{r}\frac{1}{{(\sqrt{-\frac{p}{y}}+1)}^2}=c_{1}y\end{array}$$

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

$$\begin{array}{r}p=0\end{array}$$

Solving ODE (2)

In canonical form, the ODE is

$$\begin{array}{rl}{p}^{\prime }& =F(y,p)\\ \text{(1)}& & =-\frac{\sqrt{-py}}{y}\end{array}$$

An ode of the form $p^{\prime }=\frac{M(y,p)}{N(y,p)}$ 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 $M=-\sqrt{-py}$ 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

$$\frac{\mathrm{d}p}{\mathrm{d}y}=\frac{\mathrm{d}u}{\mathrm{d}y}y+u$$

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

$$\begin{array}{rl}\frac{\mathrm{d}u}{\mathrm{d}y}y+u& =-\sqrt{-u}\\ \frac{\mathrm{d}u}{\mathrm{d}y}& =\frac{-\sqrt{-u\left(y\right)}-u\left(y\right)}{y}\end{array}$$

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

$$\begin{array}{}\text{(2)}& u^{\prime }\left(y\right)=-\frac{\sqrt{-u\left(y\right)}+u\left(y\right)}{y}\end{array}$$

is separable as it can be written as

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

Where

$$\begin{array}{rl}f(y)& =\frac{1}{y}\\ g(u)& =-\sqrt{-u}-u\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(y)dy\\ \int \frac{1}{-\sqrt{-u}-u}du& =\int \frac{1}{y}dy\end{array}$$

$$-2\ln (\sqrt{-u\left(y\right)}-1)=\ln \left(y\right)+c_2$$

Taking the exponential of both sides the solution becomes

$$\frac{1}{{(\sqrt{-u\left(y\right)}-1)}^2}=c_{2}y$$

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

$$\begin{array}{rl}u\left(y\right)& =-1\\ u\left(y\right)& =0\end{array}$$

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

$$\begin{array}{rl}\frac{1}{{(\sqrt{-u\left(y\right)}-1)}^2}& =c_{2}y\\ u\left(y\right)& =-1\\ u\left(y\right)& =0\end{array}$$

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

$$\begin{array}{r}\frac{1}{{(\sqrt{-\frac{p}{y}}-1)}^2}=c_{2}y\end{array}$$

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

$$\begin{array}{r}p=-y\end{array}$$

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

$$\begin{array}{r}p=0\end{array}$$

For solution (1) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{1}{{(\sqrt{-\frac{y^{\prime }}{y}}+1)}^2}=c_{1}y\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& y^{\prime }& =\frac{-\frac{2c_{1}y(\sqrt{c_{1}y}-1)}{\sqrt{c_{1}y}}+c_{1}y-1}{c_1}\text{(2)}& y^{\prime }& =\frac{-\frac{2c_{1}y(\sqrt{c_{1}y}+1)}{\sqrt{c_{1}y}}+c_{1}y-1}{c_1}\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

$$\begin{array}{rl}\int -\frac{\sqrt{c_{1}y}{c}_1}{c_{1}y\sqrt{c_{1}y}-2c_{1}y+\sqrt{c_{1}y}}dy& =dx\\ \frac{2+(-2\sqrt{c_{1}y}+2)\ln (\sqrt{c_{1}y}-1)}{\sqrt{c_{1}y}-1}& =x+c_3\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{c_{1}y\sqrt{c_{1}y}-2c_{1}y+\sqrt{c_{1}y}}{\sqrt{c_{1}y}{c}_1}& =0\end{array}$$

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.

$$\begin{array}{r}y=\frac{1}{c_1}\end{array}$$

Solving Eq. (2)

Integrating gives

$$\begin{array}{rl}\int -\frac{\sqrt{c_{1}y}{c}_1}{c_{1}y\sqrt{c_{1}y}+2c_{1}y+\sqrt{c_{1}y}}dy& =dx\\ \frac{-2+(-2\sqrt{c_{1}y}-2)\ln (\sqrt{c_{1}y}+1)}{\sqrt{c_{1}y}+1}& =x+c_4\end{array}$$

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

$$\begin{array}{r}{y}^{\prime }=0\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int 0dx+c_5\\ y& =c_5\end{array}$$

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

$$\begin{array}{r}\frac{1}{{(\sqrt{-\frac{y^{\prime }}{y}}-1)}^2}=c_{2}y\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& y^{\prime }& =-\frac{\frac{2c_{2}y(\sqrt{c_{2}y}+1)}{\sqrt{c_{2}y}}-c_{2}y+1}{c_2}\text{(2)}& y^{\prime }& =-\frac{\frac{2c_{2}y(\sqrt{c_{2}y}-1)}{\sqrt{c_{2}y}}-c_{2}y+1}{c_2}\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

$$\begin{array}{rl}\int -\frac{\sqrt{c_{2}y}{c}_2}{c_{2}y\sqrt{c_{2}y}+2c_{2}y+\sqrt{c_{2}y}}dy& =dx\\ \frac{-2+(-2\sqrt{c_{2}y}-2)\ln (\sqrt{c_{2}y}+1)}{\sqrt{c_{2}y}+1}& =x+c_6\end{array}$$

Solving Eq. (2)

Integrating gives

$$\begin{array}{rl}\int -\frac{\sqrt{c_{2}y}{c}_2}{c_{2}y\sqrt{c_{2}y}-2c_{2}y+\sqrt{c_{2}y}}dy& =dx\\ \frac{2+(-2\sqrt{c_{2}y}+2)\ln (\sqrt{c_{2}y}-1)}{\sqrt{c_{2}y}-1}& =x+c_7\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{c_{2}y\sqrt{c_{2}y}-2c_{2}y+\sqrt{c_{2}y}}{\sqrt{c_{2}y}{c}_2}& =0\end{array}$$

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.

$$\begin{array}{r}y=\frac{1}{c_2}\end{array}$$

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

$$\begin{array}{r}{y}^{\prime }=-y\end{array}$$

Integrating gives

$$\begin{array}{rl}\int -\frac{1}{y}dy& =dx\\ -\ln \left(y\right)& =x+c_8\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-y& =0\end{array}$$

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.

$$\begin{array}{r}y=0\end{array}$$

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

$$\begin{array}{r}{y}^{\prime }=0\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int 0dx+c_9\\ y& =c_9\end{array}$$

Will add steps showing solving for IC soon.

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

$$\begin{array}{rl}y& =0\\ y& =c_5\\ y& =c_9\\ y& =\frac{1}{c_1}\\ y& =\frac{\mathrm{LambertW}{(-{\mathrm{e}}^{\frac{c_6}{2}+\frac{x}{2}})}^2+2\mathrm{LambertW}(-{\mathrm{e}}^{\frac{c_6}{2}+\frac{x}{2}})+1}{\mathrm{LambertW}{(-{\mathrm{e}}^{\frac{c_6}{2}+\frac{x}{2}})}^2{c}_2}\\ y& =\frac{\mathrm{LambertW}{\left({\mathrm{e}}^{\frac{c_3}{2}+\frac{x}{2}}\right)}^2+2\mathrm{LambertW}\left({\mathrm{e}}^{\frac{c_3}{2}+\frac{x}{2}}\right)+1}{\mathrm{LambertW}{\left({\mathrm{e}}^{\frac{c_3}{2}+\frac{x}{2}}\right)}^2{c}_1}\\ y& ={\mathrm{e}}^{-x-c_8}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =0\\ y& =c_5\\ y& =c_9\\ y& =\frac{1}{c_1}\\ y& =\frac{\mathrm{LambertW}{(-{\mathrm{e}}^{\frac{c_6}{2}+\frac{x}{2}})}^2+2\mathrm{LambertW}(-{\mathrm{e}}^{\frac{c_6}{2}+\frac{x}{2}})+1}{\mathrm{LambertW}{(-{\mathrm{e}}^{\frac{c_6}{2}+\frac{x}{2}})}^2{c}_2}\\ y& =\frac{\mathrm{LambertW}{\left({\mathrm{e}}^{\frac{c_3}{2}+\frac{x}{2}}\right)}^2+2\mathrm{LambertW}\left({\mathrm{e}}^{\frac{c_3}{2}+\frac{x}{2}}\right)+1}{\mathrm{LambertW}{\left({\mathrm{e}}^{\frac{c_3}{2}+\frac{x}{2}}\right)}^2{c}_1}\\ y& ={\mathrm{e}}^{-x-c_8}\end{array}$$

✓ Maple. Time used: 1.138 (sec). Leaf size: 178

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

$$\begin{array}{rl}y& =c_1\\ y& =0\\ y& ={\mathrm{e}}^{-\int {\mathrm{e}}^{2\mathrm{RootOf}({\mathrm{e}}^{\mathit{\text{\_Z}}}\ln \left({({\mathrm{e}}^{\mathit{\text{\_Z}}}-1)}^2\right)+c_1{\mathrm{e}}^{\mathit{\text{\_Z}}}-2{\mathrm{e}}^{\mathit{\text{\_Z}}}\mathit{\text{\_Z}}+x{\mathrm{e}}^{\mathit{\text{\_Z}}}-\ln \left({({\mathrm{e}}^{\mathit{\text{\_Z}}}-1)}^2\right)-c_1+2\mathit{\text{\_Z}}-x+2)}dx+2\int {\mathrm{e}}^{\mathrm{RootOf}({\mathrm{e}}^{\mathit{\text{\_Z}}}\ln \left({({\mathrm{e}}^{\mathit{\text{\_Z}}}-1)}^2\right)+c_1{\mathrm{e}}^{\mathit{\text{\_Z}}}-2{\mathrm{e}}^{\mathit{\text{\_Z}}}\mathit{\text{\_Z}}+x{\mathrm{e}}^{\mathit{\text{\_Z}}}-\ln \left({({\mathrm{e}}^{\mathit{\text{\_Z}}}-1)}^2\right)-c_1+2\mathit{\text{\_Z}}-x+2)}dx-x+c_2}\\ y& =\frac{c_2{(\mathrm{LambertW}\left(c_1{\mathrm{e}}^{-1+\frac{x}{2}}\right)+1)}^2}{\mathrm{LambertW}{\left(c_1{\mathrm{e}}^{-1+\frac{x}{2}}\right)}^2}\\ y& =\frac{c_2{(\mathrm{LambertW}(-c_1{\mathrm{e}}^{-1+\frac{x}{2}})+1)}^2}{\mathrm{LambertW}{(-c_1{\mathrm{e}}^{-1+\frac{x}{2}})}^2}\end{array}$$

Maple trace

Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve e\ 
ach resulting ODE. 
      *** Sublevel 3 *** 
      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 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = -(-_b(_a))^(3/2)-_b(_a 
)^2, _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      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 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = (-_b(_a))^(3/2)-_b(_a) 
^2, _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
-> Calling odsolve with the ODE, diff(y(x),x) = 0, y(x), singsol = none 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful
 

✓ Mathematica. Time used: 1.974 (sec). Leaf size: 361

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

$$\begin{array}{r}y(x)\to \text{InverseFunction}[-4(\frac{1}{2}\log (2\sqrt{\mathrm{\#}\text{1}}-i{c}_1)-\frac{i{c}_1}{2(2\sqrt{\mathrm{\#}\text{1}}-i{c}_1)})\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[-4(\frac{i{c}_1}{2(2\sqrt{\mathrm{\#}\text{1}}+i{c}_1)}+\frac{1}{2}\log (2\sqrt{\mathrm{\#}\text{1}}+i{c}_1))\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[-4(\frac{1}{2}\log (2\sqrt{\mathrm{\#}\text{1}}-i(-c_1))-\frac{i(-c_1)}{2(2\sqrt{\mathrm{\#}\text{1}}-i(-c_1))})\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[-4(\frac{i(-c_1)}{2(2\sqrt{\mathrm{\#}\text{1}}+i(-1)c_1)}+\frac{1}{2}\log (2\sqrt{\mathrm{\#}\text{1}}+i(-1)c_1))\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[-4(\frac{1}{2}\log (2\sqrt{\mathrm{\#}\text{1}}-i{c}_1)-\frac{i{c}_1}{2(2\sqrt{\mathrm{\#}\text{1}}-i{c}_1)})\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[-4(\frac{i{c}_1}{2(2\sqrt{\mathrm{\#}\text{1}}+i{c}_1)}+\frac{1}{2}\log (2\sqrt{\mathrm{\#}\text{1}}+i{c}_1))\mathrm{\&}][x+c_2]\end{array}$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)*Derivative(y(x), (x, 2))**2 + Derivative(y(x), x)**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE (-y(x)*Derivative(y(x), (x, 2))**2)**(1/3)/2 - sqrt(3)*I*(-y(x)*Derivative(y(x), (x, 2))**2)**(1/3)/2 + Derivative(y(x), x) cannot be solved by the factorable group method