2.1.10 Problem 10

Internal problem ID [9080]
Book : Second order enumerated odes
Section : section 1
Problem number : 10
Date solved : Friday, April 25, 2025 at 05:40:21 PM
CAS classification : [[_2nd_order, _quadrature]]

Solved as second order missing y ode

Time used: 0.478 (sec)

Solve

$$\begin{array}{rl}{y^{\prime \prime }}^2& =x\end{array}$$

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(x)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(x)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{u^{\prime }\left(x\right)}^2-x=0\end{array}$$

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

Solved using first_order_ode_parametric method

Time used: 0.109 (sec)

Let $u^{\prime }$ be a parameter $\lambda$. The ode becomes

$$\begin{array}{r}{\lambda }^2-x=0\end{array}$$

Isolating $x$ gives

$$\begin{array}{r}x={\lambda }^2\\ x=F(u,\lambda )\end{array}$$

Now we generate an ode in $u\left(\lambda \right)$ using

$$\begin{array}{rl}\frac{d}{d\lambda }u\left(\lambda \right)& =\frac{\lambda \frac{\partial F}{\partial \lambda }}{1-\frac{\partial F}{\partial u}}\\ & =2{\lambda }^2\\ & =2{\lambda }^2\end{array}$$

Which is now solved for $u$.

Since the ode has the form $u^{\prime }=f(\lambda )$, then we only need to integrate $f(\lambda )$.

$$\begin{array}{rl}\int du& =\int 2{\lambda }^{2}d\lambda \\ u& =\frac{2{\lambda }^3}{3}+c_1\end{array}$$

Now that we have found solution $u$, we have two equations with parameter $\lambda$. They are

$$\begin{array}{rl}u& =\frac{2{\lambda }^3}{3}+c_1\\ x& ={\lambda }^2\end{array}$$

Eliminating $\lambda$ gives the solution for $u$. Solving for $u\left(x\right)$ gives

$$\begin{array}{rl}u\left(x\right)& =c_1-\frac{2x^{3/2}}{3}\\ u\left(x\right)& =c_1+\frac{2x^{3/2}}{3}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}u\left(x\right)& =c_1-\frac{2x^{3/2}}{3}\\ u\left(x\right)& =c_1+\frac{2x^{3/2}}{3}\end{array}$$

In summary, these are the solution found for $y\left(x\right)$

$$\begin{array}{rl}u\left(x\right)& =c_1-\frac{2x^{3/2}}{3}\\ u\left(x\right)& =c_1+\frac{2x^{3/2}}{3}\end{array}$$

For solution $u\left(x\right)=c_1-\frac{2x^{3/2}}{3}$, since $u=\frac{d}{dx}y\left(x\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{d}{dx}y\left(x\right)=c_1-\frac{2x^{3/2}}{3}\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int c_1-\frac{2x^{3/2}}{3}dx\\ y& =c_{1}x-\frac{4x^{5/2}}{15}+c_2\end{array}$$

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

$$\begin{array}{rl}y& =c_{1}x-\frac{4x^{5/2}}{15}+c_2\end{array}$$

For solution $u\left(x\right)=c_1+\frac{2x^{3/2}}{3}$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=c_1+\frac{2x^{3/2}}{3}\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int c_1+\frac{2x^{3/2}}{3}dx\\ y& =c_{1}x+\frac{4x^{5/2}}{15}+c_3\end{array}$$

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

$$\begin{array}{rl}y& =c_{1}x+\frac{4x^{5/2}}{15}+c_3\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =c_{1}x-\frac{4x^{5/2}}{15}+c_2\\ y& =c_{1}x+\frac{4x^{5/2}}{15}+c_3\end{array}$$

✓ Maple. Time used: 0.018 (sec). Leaf size: 27

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

$$\begin{array}{rl}y& =\frac{4x^{5/2}}{15}+c_{1}x+c_2\\ y& =-\frac{4x^{5/2}}{15}+c_{1}x+c_2\end{array}$$

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
 

✓ Mathematica. Time used: 0.007 (sec). Leaf size: 41

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

$$\begin{array}{r}y(x)\to -\frac{4x^{5/2}}{15}+c_{2}x+c_1\\ y(x)\to \frac{4x^{5/2}}{15}+c_{2}x+c_1\end{array}$$

✓ Sympy. Time used: 0.373 (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)
 

$$[y\left(x\right)=C_1+C_{2}x-\frac{4x^{\frac{5}{2}}}{15},y\left(x\right)=C_1+C_{2}x+\frac{4x^{\frac{5}{2}}}{15}]$$