2.1.10 Problem 10

Internal problem ID [9081]
Book : Second order enumerated odes
Section : section 1
Problem number : 10
Date solved : Monday, January 27, 2025 at 05:32:13 PM
CAS classification : [[_2nd_order, _quadrature]]

Solve

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

Solved as second order missing y ode

Time used: 0.244 (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}-x = 0 \end{align*}

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

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

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

\begin{align*} \int {du} &= \int {\sqrt {x}\, dx}\\ u \left (x \right ) &= \frac {2 x^{{3}/{2}}}{3} + c_1 \end{align*}

Solving Eq. (2)

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

\begin{align*} \int {du} &= \int {-\sqrt {x}\, dx}\\ u \left (x \right ) &= -\frac {2 x^{{3}/{2}}}{3} + c_2 \end{align*}

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

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

For solution \(u \left (x \right ) = -\frac {2 x^{{3}/{2}}}{3}+c_2\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

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

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

\begin{align*} y^{\prime } = \frac {2 x^{{3}/{2}}}{3}+c_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 {\frac {2 x^{{3}/{2}}}{3}+c_1\, dx}\\ y &= c_1 x +\frac {4 x^{{5}/{2}}}{15} + c_4 \end{align*}

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

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

Maple dsolve solution

Solving time : 0.023 (sec)
Leaf size : 27

dsolve(diff(diff(y(x),x),x)^2 = x,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*}

Mathematica DSolve solution

Solving time : 0.007 (sec)
Leaf size : 41

DSolve[{(D[y[x],{x,2}])^2==x,{}},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*}