1.19 problem 19

Internal problem ID [8056]
Book : Second order enumerated odes
Section : section 1
Problem number : 19
Date solved : Monday, October 21, 2024 at 04:44:56 PM
CAS classification : [[_2nd_order, _missing_y]]

Solve

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

1.19.1 Solved as second order missing y ode

Time used: 0.474 (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*} p \left (x \right ) &= -p^{2}+x\tag {1A} \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*}

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 \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

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

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

Will add steps showing solving for IC soon.

1.19.2 Maple step by step solution

1.19.3 Maple trace

Methods for second order ODEs:
 

1.19.4 Maple dsolve solution

Solving time : 0.034 (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*}

1.19.5 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*}