2.1.19 problem 19

Internal problem ID [8502]
Book : Second order enumerated odes
Section : section 1
Problem number : 19
Date solved : Sunday, November 10, 2024 at 03:55:13 AM
CAS classification : [[_2nd_order, _missing_y]]

Solve

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

Solved as second order missing y ode

Time used: 0.503 (sec)

This is second order ode with missing dependent variable \(y\). Let

\begin{align*} p(x) &= y^{\prime } \end{align*}

Then

\begin{align*} p'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} {p^{\prime }\left (x \right )}^{2}+p \left (x \right )-x = 0 \end{align*}

Which is now solve for \(p(x)\) as first order ode. Let \(p=p^{\prime }\left (x \right )\) the ode becomes

\begin{align*} p^{2}+p -x = 0 \end{align*}

Solving for \(p \left (x \right )\) from the above results in

\begin{align*} \tag{1} p \left (x \right ) &= -p^{2}+x \\ \end{align*}

This has the form

\begin{align*} p=xf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=p'(x)\). The above ode is dAlembert ode which is now solved.

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

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(p \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= 1\\ g &= -p^{2} \end{align*}

Hence (2) becomes

\begin{align*} p -1 = -2 p p^{\prime }\left (x \right )\tag {2A} \end{align*}

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

\begin{align*} p^{\prime }\left (x \right ) = -\frac {p \left (x \right )-1}{2 p \left (x \right )}\tag {3} \end{align*}

Inverting the above ode gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = -\frac {2 p}{p -1}\tag {4} \end{align*}

This ODE is now solved for \(x \left (p \right )\). Integrating gives

\begin{align*} x \left (p \right )&=\int -\frac {2 p}{p -1}d p +c_1\\ x \left (p \right )&=-2 p -2 \ln \left (p -1\right )+c_1\tag {5} \end{align*}

Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into (1A).

Eliminating \(p\) from the following two equations

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

results in

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

Substituting the above into Eq (1A) and simplifying gives

\begin{align*} p \left (x \right ) &= -\operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{2}-2 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )+x -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 } = -\operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{2}-2 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )+x -1 \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 {-\operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{2}-2 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )+x -1\, dx}\\ y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )+\frac {x^{2}}{2}-x + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-\frac {x}{2}-1+\frac {c_1}{2}}\right )+\frac {x^{2}}{2}-x +c_2 \\ \end{align*}

Maple step by step solution

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) = (-_b(_a)+_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]
 

Maple dsolve solution

Solving time : 0.113 (sec)
Leaf size : 122

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

\begin{align*} y &= \int \left (-{\mathrm e}^{2 \operatorname {RootOf}\left (\textit {\_Z} -x -2 \,{\mathrm e}^{\textit {\_Z}}+2+c_{1} -\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z} -x -2 \,{\mathrm e}^{\textit {\_Z}}+2+c_{1} -\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}+x \right )d x -x +c_{2} \\ y &= \frac {2 \operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-\frac {x}{2}-1}\right )^{3}}{3}+3 \operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-\frac {x}{2}-1}\right )^{2}+4 \operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-\frac {x}{2}-1}\right )+\frac {x^{2}}{2}-x +c_{2} \\ \end{align*}

Mathematica DSolve solution

Solving time : 16.306 (sec)
Leaf size : 172

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

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