2.1.19 Problem 19

Internal problem ID [9090]
Book : Second order enumerated odes
Section : section 1
Problem number : 19
Date solved : Sunday, March 30, 2025 at 02:06:45 PM
CAS classification : [[_2nd_order, _missing_y]]

Solved as second order missing y ode

Time used: 0.969 (sec)

Solve

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

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(x)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(x)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{u^{\prime }\left(x\right)}^2+u\left(x\right)-x=0\end{array}$$

Which is now solved for $u(x)$ as first order ode.

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

$$\begin{array}{r}{p}^2+u-x=0\end{array}$$

Solving for $u$ from the above results in

$$\begin{array}{}\text{(1)}& u& =-p^2+x\end{array}$$

This has the form

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

Where $f,g$ are functions of $p=u^{\prime }(x)$. 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 $u=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =1\\ g& =-p^2\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p-1=-2p{p}^{\prime }\left(x\right)\end{array}$$

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

$$\begin{array}{r}p-1=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =1\end{array}$$

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

$$\begin{array}{r}u=x-1\end{array}$$

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

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

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

Integrating gives

$$\begin{array}{rl}\int -\frac{2p}{p-1}dp& =dx\\ -2p-2\ln (p-1)& =x+c_1\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{p-1}{2p}& =0\end{array}$$

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.

$$\begin{array}{r}p\left(x\right)=1\end{array}$$

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

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

In summary, these are the solution found for $u(x)$

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

For solution $u=x-1$, since $u=y^{\prime }\left(x\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }\left(x\right)=x-1\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int x-1dx\\ y& =\frac{1}{2}{x}^2-x+c_2\end{array}$$

For solution $u\left(x\right)=-{({\mathrm{e}}^{-\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-1-\frac{x}{2}-\frac{c_1}{2}}+1)}^2+x$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

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

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

$$\begin{array}{rl}y& =\int (-{\mathrm{e}}^{-2\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-2-x-c_1}-2{\mathrm{e}}^{-\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-1-\frac{x}{2}-\frac{c_1}{2}}+x-1)dx+c_3\end{array}$$

For solution $u\left(x\right)=x-1$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=x-1\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int x-1dx\\ y& =\frac{1}{2}{x}^2-x+c_4\end{array}$$

In summary, these are the solution found for $(y)$

$$\begin{array}{rl}y& =\frac{1}{2}{x}^2-x+c_2\\ y& =\int (-{\mathrm{e}}^{-2\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-2-x-c_1}-2{\mathrm{e}}^{-\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-1-\frac{x}{2}-\frac{c_1}{2}}+x-1)dx+c_3\\ y& =\frac{1}{2}{x}^2-x+c_4\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\int (-{\mathrm{e}}^{-2\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-2-x-c_1}-2{\mathrm{e}}^{-\mathrm{LambertW}\left({\mathrm{e}}^{-1-\frac{x}{2}-\frac{c_1}{2}}\right)-1-\frac{x}{2}-\frac{c_1}{2}}+x-1)dx+c_3\\ y& =\frac{1}{2}{x}^2-x+c_2\\ y& =\frac{1}{2}{x}^2-x+c_4\end{array}$$

✓ Maple. Time used: 0.334 (sec). Leaf size: 122

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

$$\begin{array}{rl}y& =\int (-{\mathrm{e}}^{2\mathrm{RootOf}(\mathit{\text{\_Z}}-x-2{\mathrm{e}}^{\mathit{\text{\_Z}}}+2+c_1-\ln \left({\mathrm{e}}^{\mathit{\text{\_Z}}}{({\mathrm{e}}^{\mathit{\text{\_Z}}}-2)}^2\right))}+2{\mathrm{e}}^{\mathrm{RootOf}(\mathit{\text{\_Z}}-x-2{\mathrm{e}}^{\mathit{\text{\_Z}}}+2+c_1-\ln \left({\mathrm{e}}^{\mathit{\text{\_Z}}}{({\mathrm{e}}^{\mathit{\text{\_Z}}}-2)}^2\right))}+x)dx-x+c_2\\ y& =\frac{2\mathrm{LambertW}{(-c_1{\mathrm{e}}^{-1-\frac{x}{2}})}^3}{3}+3\mathrm{LambertW}{(-c_1{\mathrm{e}}^{-1-\frac{x}{2}})}^2+4\mathrm{LambertW}(-c_1{\mathrm{e}}^{-1-\frac{x}{2}})+\frac{x^2}{2}-x+c_2\end{array}$$

Maple trace

Methods for second order ODEs: 
Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each\ 
 resulting ODE. 
   *** Sublevel 2 *** 
   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 
   -> Calling odsolve with the ODE, diff(_b(_a),_a) = (_a-_b(_a))^(1/2), _b(_a) 
, HINT = [[1, 1]] 
      *** Sublevel 3 *** 
      symmetry methods on request 
      1st order, trying reduction of order with given symmetries: 
[1, 1] 
      1st order, trying the canonical coordinates of the invariance group 
         -> Calling odsolve with the ODE, diff(y(x),x) = 1, y(x) 
            *** Sublevel 4 *** 
            Methods for first order ODEs: 
            --- Trying classification methods --- 
            trying a quadrature 
            trying 1st order linear 
            <- 1st order linear successful 
      <- 1st order, canonical coordinates successful 
   <- differential order: 2; canonical coordinates successful 
   <- differential order 2; missing variables successful 
------------------- 
* Tackling next ODE. 
   *** Sublevel 2 *** 
   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 
   -> Calling odsolve with the ODE, diff(_b(_a),_a) = -(_a-_b(_a))^(1/2), _b(_a 
), HINT = [[1, 1]] 
      *** Sublevel 3 *** 
      symmetry methods on request 
      1st order, trying reduction of order with given symmetries: 
[1, 1] 
      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
 

✓ Mathematica. Time used: 17.332 (sec). Leaf size: 172

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

$$\begin{array}{r}y(x)\to \frac{2}{3}W\left(e^{-\frac{x}{2}-1-\frac{c_1}{2}}\right){}^3+3W\left(e^{-\frac{x}{2}-1-\frac{c_1}{2}}\right){}^2+4W\left(e^{-\frac{x}{2}-1-\frac{c_1}{2}}\right)+\frac{x^2}{2}-x+c_2\\ y(x)\to \frac{2}{3}W(-e^{\frac{1}{2}(-x-2+c_1)}){}^3+3W(-e^{\frac{1}{2}(-x-2+c_1)}){}^2+4W(-e^{\frac{1}{2}(-x-2+c_1)})+\frac{x^2}{2}-x+c_2\\ y(x)\to \frac{x^2}{2}-x+c_2\end{array}$$

✗ Sympy

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

NotImplementedError : multiple generators [log(sqrt(-_X0 + x) + 1), sqrt(-_X0 + x)] 
No algorithms are implemented to solve equation C1 + x - 2*sqrt(-_X0 + x) + 2*log(sqrt(-_X0 + x) + 1)