2.1.23 Problem 23

Internal problem ID [9094]
Book : Second order enumerated odes
Section : section 1
Problem number : 23
Date solved : Sunday, March 30, 2025 at 02:06:52 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order missing x ode

Time used: 1.880 (sec)

Solve

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

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

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

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

In canonical form, the ODE is

$$\begin{array}{rl}{p}^{\prime }& =F(y,p)\\ & =-\frac{p^2+y}{p}\end{array}$$

This is a Bernoulli ODE.

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

The standard Bernoulli ODE has the form

$$\begin{array}{}\text{(2)}& p^{\prime }=f_0(y)p+f_1(y)p^n\end{array}$$

Comparing this to (1) shows that

$$\begin{array}{rl}{f}_0& =-1\\ f_1& =-y\end{array}$$

The first step is to divide the above equation by $p^n$ which gives

$$\begin{array}{}\text{(3)}& \frac{p^{\prime }}{p^n}=f_0(y)p^{1-n}+f_1(y)\end{array}$$

The next step is use the substitution $v=p^{1-n}$ in equation (3) which generates a new ODE in $v\left(y\right)$ which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution $p(y)$ which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

$$\begin{array}{rl}{f}_0(y)& =-1\\ f_1(y)& =-y\\ n& =-1\end{array}$$

Dividing both sides of ODE (1) by $p^n=\frac{1}{p}$ gives

$$\begin{array}{}\text{(4)}& p^{\prime }p& =-p^2-y\end{array}$$

Let

$$\begin{array}{rl}v& =p^{1-n}\\ \text{(5)}& & =p^2\end{array}$$

Taking derivative of equation (5) w.r.t $y$ gives

$$\begin{array}{}\text{(6)}& v^{\prime }& =2p{p}^{\prime }\end{array}$$

Substituting equations (5) and (6) into equation (4) gives

$$\begin{array}{rl}\frac{v^{\prime }\left(y\right)}{2}& =-v\left(y\right)-y\\ \text{(7)}& v^{\prime }& =-2v-2y\end{array}$$

The above now is a linear ODE in $v\left(y\right)$ which is now solved.

In canonical form a linear first order is

$$\begin{array}{rl}{v}^{\prime }\left(y\right)+q(y)v\left(y\right)& =p(y)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(y)& =2\\ p(y)& =-2y\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdy}\\ & ={\mathrm{e}}^{\int 2dy}\\ & ={\mathrm{e}}^{2y}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}y}\left(\mu v\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}y}\left(\mu v\right)& =\left(\mu \right)(-2y)\\ \frac{\mathrm{d}}{\mathrm{d}y}\left(v{\mathrm{e}}^{2y}\right)& =\left({\mathrm{e}}^{2y}\right)(-2y)\\ \mathrm{d}\left(v{\mathrm{e}}^{2y}\right)& =(-2y{\mathrm{e}}^{2y})\mathrm{d}y\end{array}$$

Integrating gives

$$\begin{array}{rl}v{\mathrm{e}}^{2y}& =\int -2y{\mathrm{e}}^{2y}dy\\ & =-\frac{(2y-1){\mathrm{e}}^{2y}}{2}+c_1\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^{2y}$ gives the final solution

$$v\left(y\right)=-y+\frac{1}{2}+{\mathrm{e}}^{-2y}{c}_1$$

The substitution $v=p^{1-n}$ is now used to convert the above solution back to $p$ which results in

$$p^2=-y+\frac{1}{2}+{\mathrm{e}}^{-2y}{c}_1$$

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

$$\begin{array}{r}{y^{\prime }}^2=-y+\frac{1}{2}+{\mathrm{e}}^{-2y}{c}_1\end{array}$$

Let $p=y^{\prime }$ the ode becomes

$$\begin{array}{r}{p}^2=-y+\frac{1}{2}+{\mathrm{e}}^{-2y}{c}_1\end{array}$$

Solving for $y$ from the above results in

$$\begin{array}{}\text{(1)}& y& =\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p^2-1}\right)}{2}-p^2+\frac{1}{2}\end{array}$$

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p^2-1}\right)}{2}-p^2+\frac{1}{2}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=(\frac{2p\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p^2-1}\right)}{1+\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p^2-1}\right)}-2p)p^{\prime }\left(x\right)\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

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

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{-1}\right)}{2}+\frac{1}{2}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=\frac{p\left(x\right)}{\frac{2p\left(x\right)\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p{\left(x\right)}^2-1}\right)}{1+\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p{\left(x\right)}^2-1}\right)}-2p\left(x\right)}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^{p\left(x\right)}\frac{1}{-\frac{1}{2}-\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2{\tau }^2-1}\right)}{2}}d\tau =x+c_2$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{1}{2}-\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{2p^2-1}\right)}{2}& =0\end{array}$$

for $p\left(x\right)$. 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.

$$\begin{array}{r}p\left(x\right)=-\frac{\sqrt{2}\sqrt{\ln (-\frac{1}{2c_1})}}{2}\\ p\left(x\right)=\frac{\sqrt{2}\sqrt{\ln (-\frac{1}{2c_1})}}{2}\end{array}$$

Substituing the above solution for $p$ in (2A) gives

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

Will add steps showing solving for IC soon.

The solution

$$y=\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{-1}\right)}{2}+\frac{1}{2}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$y=\frac{\mathrm{LambertW}\left(2c_1{\mathrm{e}}^{\ln (-\frac{1}{2c_1})-1}\right)}{2}-\frac{\ln (-\frac{1}{2c_1})}{2}+\frac{1}{2}$$

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

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

✓ Maple. Time used: 0.021 (sec). Leaf size: 61

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

$$\begin{array}{rl}-2{\int }_{}^y\frac{1}{\sqrt{2+4{\mathrm{e}}^{-2\mathit{\text{\_a}}}{c}_1-4\mathit{\text{\_a}}}}d\mathit{\text{\_a}}-x-c_2& =0\\ 2{\int }_{}^y\frac{1}{\sqrt{2+4{\mathrm{e}}^{-2\mathit{\text{\_a}}}{c}_1-4\mathit{\text{\_a}}}}d\mathit{\text{\_a}}-x-c_2& =0\end{array}$$

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 
   -> Computing symmetries using: way = 3 
   -> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+_b(_a)^2+_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
 

✓ Mathematica. Time used: 0.823 (sec). Leaf size: 272

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

$$\begin{array}{r}y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}-\frac{\sqrt{2}}{\sqrt{2e^{-2K[1]}{c}_1-2K[1]+1}}dK[1]\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}\frac{\sqrt{2}}{\sqrt{2e^{-2K[2]}{c}_1-2K[2]+1}}dK[2]\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}-\frac{\sqrt{2}}{\sqrt{2e^{-2K[1]}(-c_1)-2K[1]+1}}dK[1]\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}-\frac{\sqrt{2}}{\sqrt{2e^{-2K[1]}{c}_1-2K[1]+1}}dK[1]\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}\frac{\sqrt{2}}{\sqrt{2e^{-2K[2]}(-c_1)-2K[2]+1}}dK[2]\mathrm{\&}][x+c_2]\\ y(x)\to \text{InverseFunction}[{\int }_1^{\mathrm{\#}\text{1}}\frac{\sqrt{2}}{\sqrt{2e^{-2K[2]}{c}_1-2K[2]+1}}dK[2]\mathrm{\&}][x+c_2]\end{array}$$

✗ Sympy

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

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