2.1.13 Problem 13

Internal problem ID [9084]
Book : Second order enumerated odes
Section : section 1
Problem number : 13
Date solved : Wednesday, March 05, 2025 at 07:19:21 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} {y^{\prime \prime }}^{2}+y^{\prime }&=0 \end{align*}

Solved as second order missing x ode

Time used: 1.829 (sec)

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{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end{align*}

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

Factoring the ode gives these factors

\begin{align*} \tag{1} p &= 0 \\ \tag{2} {p^{\prime }}^{2} p+1 &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(p\) from

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

Solving gives \(p = 0\)

Solving equation (2)

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} p^{\prime }&=-\frac {1}{\sqrt {-p}} \\ \tag{2} p^{\prime }&=\frac {1}{\sqrt {-p}} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int -\sqrt {-p}d p &= dy\\ \frac {2 \left (-p \right )^{{3}/{2}}}{3}&= y +c_1 \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int \sqrt {-p}d p &= dy\\ -\frac {2 \left (-p \right )^{{3}/{2}}}{3}&= y +c_2 \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*} -\frac {2 \left (-y^{\prime }\right )^{{3}/{2}}}{3} = y+c_2 \end{align*}

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

\begin{align*} -\frac {2 \left (-p \right )^{{3}/{2}}}{3} = y +c_2 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= -c_2 -\frac {2 \left (-p \right )^{{3}/{2}}}{3} \\ \end{align*}

This has the form

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

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

\begin{align*} f &= 0\\ g &= -c_2 -\frac {2 \left (-p \right )^{{3}/{2}}}{3} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \sqrt {-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 = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

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

\begin{align*} y = -c_2 \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 )}{\sqrt {-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 {1}{\sqrt {-p}}d p &= dx\\ 2 \sqrt {-p}&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {-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 ) = 0 \end{align*}

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

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

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

\begin{align*} \frac {2 \left (-p \right )^{{3}/{2}}}{3} = y +c_1 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= -c_1 +\frac {2 \left (-p \right )^{{3}/{2}}}{3} \\ \end{align*}

This has the form

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

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

\begin{align*} f &= 0\\ g &= -c_1 +\frac {2 \left (-p \right )^{{3}/{2}}}{3} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = -\sqrt {-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 = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

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

\begin{align*} y = -c_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 )}{\sqrt {-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 {1}{\sqrt {-p}}d p &= dx\\ -2 \sqrt {-p}&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {-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 ) = 0 \end{align*}

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

\begin{align*} y &= -c_1 +\frac {2 \left (\frac {1}{4} x^{2}+\frac {1}{2} c_4 x +\frac {1}{4} c_4^{2}\right )^{{3}/{2}}}{3} \\ y &= -c_1 \\ \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 0 \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 {0\, dx} + c_5 \\ y &= c_5 \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = -c_1 +\frac {2 \left (\frac {1}{4} x^{2}+\frac {1}{2} c_4 x +\frac {1}{4} c_4^{2}\right )^{{3}/{2}}}{3} \]

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

Summary of solutions found

\begin{align*} y &= c_5 \\ y &= -c_1 \\ y &= -c_2 \\ y &= -c_2 -\frac {2 \left (\frac {1}{4} x^{2}+\frac {1}{2} c_3 x +\frac {1}{4} c_3^{2}\right )^{{3}/{2}}}{3} \\ \end{align*}

Solved as second order missing y ode

Time used: 0.405 (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 ) = 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 = 0 \end{align*}

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

\begin{align*} \tag{1} u \left (x \right ) &= -p^{2} \\ \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 &= 0\\ g &= -p^{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = -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 = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

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

\begin{align*} u \left (x \right ) = 0 \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 {1}{2}} \end{equation}

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

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

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

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

\begin{align*} u \left (x \right ) &= -\frac {\left (-x +2 c_1 \right )^{2}}{4} \\ \end{align*}

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

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= -\frac {\left (-x +2 c_1 \right )^{2}}{4} \\ \end{align*}

For solution \(u \left (x \right ) = 0\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 0 \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 {0\, dx} + c_2 \\ y &= c_2 \end{align*}

For solution \(u \left (x \right ) = -\frac {\left (-x +2 c_1 \right )^{2}}{4}\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\frac {\left (-x +2 c_1 \right )^{2}}{4} \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 {-\frac {\left (-x +2 c_1 \right )^{2}}{4}\, dx}\\ y &= \frac {\left (-x +2 c_1 \right )^{3}}{12} + c_3 \end{align*}

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

\begin{align*} y &= c_2 \\ y &= \frac {\left (-x +2 c_1 \right )^{3}}{12}+c_3 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_2 \\ y &= \frac {\left (-x +2 c_1 \right )^{3}}{12}+c_3 \\ \end{align*}

✓ Maple. Time used: 0.196 (sec). Leaf size: 27

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

\begin{align*} y &= c_{1} \\ y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-c_{1}^{2} x +c_{2} \\ \end{align*}

Maple trace

`Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      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 
      -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)+1/2, y(x)`         *** Sublevel 4 *** 
         Methods for third order ODEs: 
         --- Trying classification methods --- 
         trying a quadrature 
         <- quadrature successful 
      <- 2nd order ODE linearizable_by_differentiation successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      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 
      <- 2nd order ODE linearizable_by_differentiation successful`
 

✓ Mathematica. Time used: 0.023 (sec). Leaf size: 69

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

\begin{align*} y(x)\to -\frac {x^3}{12}-\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \\ y(x)\to -\frac {x^3}{12}+\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \\ \end{align*}

✓ Sympy. Time used: 0.792 (sec). Leaf size: 22

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

\[ y{\left (x \right )} = C_{1} - \frac {C_{2}^{2} x}{4} + \frac {C_{2} x^{2}}{4} - \frac {x^{3}}{12} \]