Problem 19

2.1.19 Problem 19

Solved as second order missing y ode
Maple step by step solution
Maple trace
Maple dsolve solution
Mathematica DSolve solution

Internal problem ID [9090]
Book : Second order enumerated odes
Section : section 1
Problem number : 19
Date solved : Monday, January 27, 2025 at 05:37:15 PM
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: 0.581 (sec)

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

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

Then

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

Hence the ode becomes

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

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

Let \(p=u^{\prime }\left (x \right )\) the ode becomes

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

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

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

This has the form

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

Where \(f,g\) are functions of \(p=u'(x)\). 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 \(u \left (x \right )=x f + g\) to (1A) shows that

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

Hence (2) becomes

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

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

\begin{align*} p -1 = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=1 \end{align*}

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

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

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

Integrating gives

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

Singular solutions are found by solving

\begin{align*} -\frac {p -1}{2 p}&= 0 \end{align*}

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{align*} p \left (x \right ) = 1 \end{align*}

Solving for \(p \left (x \right )\) gives

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

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

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

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

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

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{align*} y^{\prime } = 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 {x -1\, dx}\\ y &= \frac {1}{2} x^{2}-x + c_2 \end{align*}

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

\begin{align*} y^{\prime } = -\operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}-2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\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}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}-2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )+x -1\, dx}\\ y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )+\frac {x^{2}}{2}-x + c_3 \end{align*}

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

\begin{align*} y &= \frac {1}{2} x^{2}-x +c_2 \\ y &= \frac {2 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{3}}{3}+3 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )^{2}+4 \operatorname {LambertW}\left ({\mathrm e}^{-1-\frac {x}{2}-\frac {c_1}{2}}\right )+\frac {x^{2}}{2}-x +c_3 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

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 : 0.063 (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 : 17.332 (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*}

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