2.1.10 Problem 10

Internal problem ID [10369]
Book : Second order enumerated odes
Section : section 1
Problem number : 10
Date solved : Sunday, March 01, 2026 at 07:51:47 AM
CAS classification : [[_2nd_order, _quadrature]]

2.1.10.1 second order ode missing y

1.114 (sec)

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

Entering second order ode missing \(y\) solverThis 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}-x = 0 \end{align*}

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

Entering first order ode homog type G solverMultiplying the right side of the ode, which is \(\sqrt {x}\) by \(\frac {x}{u}\) gives

\begin{align*} u^{\prime }\left (x \right ) &= \left (\frac {x}{u}\right ) \sqrt {x}\\ &= \frac {x^{{3}/{2}}}{u}\\ &= F(x,u) \end{align*}

Since \(F \left (x , u\right )\) has \(u\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , u\right )\right )\\ &= \frac {3 x^{{3}/{2}}}{2 u}\\ f_u&= u \left (\frac {\partial }{\partial u}F \left (x , u\right )\right )\\ &= -\frac {x^{{3}/{2}}}{u}\\ \alpha &= \frac {f_x}{f_u} \\ &=-{\frac {3}{2}} \end{align*}

Since \(\alpha \) is independent of \(x,u\) then this is Homogeneous type G.

Let

\begin{align*} u&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x^{{3}/{2}}}} \end{align*}

Substituting the above back into \(F(x,u)\) gives

\begin{align*} F \left (z \right ) &=\frac {1}{z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(u\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{u x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {u \left (x \right )}{x^{{3}/{2}}}}\frac {1}{z \left (\frac {3}{2}-\frac {1}{z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 +\frac {2 \ln \left (\frac {-2 x^{{3}/{2}}+3 u \left (x \right )}{x^{{3}/{2}}}\right )}{3} = 0 \]

Multiplying the right side of the ode, which is \(-\sqrt {x}\) by \(\frac {x}{u}\) gives

\begin{align*} u^{\prime }\left (x \right ) &= \left (\frac {x}{u}\right ) -\sqrt {x}\\ &= -\frac {x^{{3}/{2}}}{u}\\ &= F(x,u) \end{align*}

Since \(F \left (x , u\right )\) has \(u\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , u\right )\right )\\ &= -\frac {3 x^{{3}/{2}}}{2 u}\\ f_u&= u \left (\frac {\partial }{\partial u}F \left (x , u\right )\right )\\ &= \frac {x^{{3}/{2}}}{u}\\ \alpha &= \frac {f_x}{f_u} \\ &=-{\frac {3}{2}} \end{align*}

Since \(\alpha \) is independent of \(x,u\) then this is Homogeneous type G.

Let

\begin{align*} u&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x^{{3}/{2}}}} \end{align*}

Substituting the above back into \(F(x,u)\) gives

\begin{align*} F \left (z \right ) &=-\frac {1}{z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(u\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{u x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_2 +\int _{}^{\frac {u \left (x \right )}{x^{{3}/{2}}}}\frac {1}{z \left (\frac {3}{2}+\frac {1}{z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_2 +\frac {2 \ln \left (\frac {2 x^{{3}/{2}}+3 u \left (x \right )}{x^{{3}/{2}}}\right )}{3} = 0 \]

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

\begin{align*} u \left (x \right ) &= \frac {{\mathrm e}^{\frac {3 c_1}{2}}}{3}+\frac {2 x^{{3}/{2}}}{3} \\ u \left (x \right ) &= \frac {{\mathrm e}^{\frac {3 c_2}{2}}}{3}-\frac {2 x^{{3}/{2}}}{3} \\ \end{align*}

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

\begin{align*} u \left (x \right ) &= \frac {{\mathrm e}^{\frac {3 c_1}{2}}}{3}+\frac {2 x^{{3}/{2}}}{3} \\ u \left (x \right ) &= \frac {{\mathrm e}^{\frac {3 c_2}{2}}}{3}-\frac {2 x^{{3}/{2}}}{3} \\ \end{align*}

For solution \(u \left (x \right ) = \frac {{\mathrm e}^{\frac {3 c_1}{2}}}{3}+\frac {2 x^{{3}/{2}}}{3}\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = \frac {{\mathrm e}^{\frac {3 c_1}{2}}}{3}+\frac {2 x^{{3}/{2}}}{3} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {\frac {{\mathrm e}^{\frac {3 c_1}{2}}}{3}+\frac {2 x^{{3}/{2}}}{3}\, dx}\\ y &= \frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_1}{2}}}{3} + c_3 \end{align*}

\begin{align*} y&= \frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_1}{2}}}{3}+c_3 \end{align*}

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

\begin{align*} y &= \frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_1}{2}}}{3}+c_3 \\ \end{align*}

For solution \(u \left (x \right ) = \frac {{\mathrm e}^{\frac {3 c_2}{2}}}{3}-\frac {2 x^{{3}/{2}}}{3}\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = \frac {{\mathrm e}^{\frac {3 c_2}{2}}}{3}-\frac {2 x^{{3}/{2}}}{3} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {\frac {{\mathrm e}^{\frac {3 c_2}{2}}}{3}-\frac {2 x^{{3}/{2}}}{3}\, dx}\\ y &= -\frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_2}{2}}}{3} + c_4 \end{align*}

\begin{align*} y&= -\frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_2}{2}}}{3}+c_4 \end{align*}

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

\begin{align*} y &= -\frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_2}{2}}}{3}+c_4 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= -\frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_2}{2}}}{3}+c_4 \\ y &= \frac {4 x^{{5}/{2}}}{15}+\frac {x \,{\mathrm e}^{\frac {3 c_1}{2}}}{3}+c_3 \\ \end{align*}

2.1.10.2 ✓ Maple. Time used: 0.007 (sec). Leaf size: 27

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

\begin{align*} y &= \frac {4 x^{{5}/{2}}}{15}+c_1 x +c_2 \\ y &= -\frac {4 x^{{5}/{2}}}{15}+c_1 x +c_2 \\ \end{align*}

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 a quadrature 
   <- quadrature successful 
------------------- 
* Tackling next ODE. 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   <- quadrature successful
 

2.1.10.3 ✓ Mathematica. Time used: 0.003 (sec). Leaf size: 41

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

\begin{align*} y(x)&\to -\frac {4 x^{5/2}}{15}+c_2 x+c_1\\ y(x)&\to \frac {4 x^{5/2}}{15}+c_2 x+c_1 \end{align*}

2.1.10.4 ✓ Sympy. Time used: 0.358 (sec). Leaf size: 31

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

\[ \left [ y{\left (x \right )} = C_{1} + C_{2} x - \frac {4 x^{\frac {5}{2}}}{15}, \ y{\left (x \right )} = C_{1} + C_{2} x + \frac {4 x^{\frac {5}{2}}}{15}\right ] \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('nth_algebraic', 'nth_algebraic_Integral')