2.1.61 Problem 61

Internal problem ID [8773]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 61
Date solved : Wednesday, March 05, 2025 at 06:47:41 AM
CAS classification : [[_2nd_order, _quadrature]]

Solve

$$\begin{array}{rl}y{y}^{\prime \prime }& =0\end{array}$$

Factoring the ode gives these factors

$$\begin{array}{}\text{(1)}& y& =0\text{(2)}& y^{\prime \prime }& =0\end{array}$$

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for $y$ from

$$\begin{array}{r}y=0\end{array}$$

Solving gives $y=0$

Solving equation (2)

Solved as second order missing x ode

Time used: 0.072 (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

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

Then

$$\begin{array}{rl}{y}^{″}& =\frac{dp}{dx}\\ & =\frac{dp}{dy}\frac{dy}{dx}\\ & =p\frac{dp}{dy}\end{array}$$

Hence the ode becomes

$$\begin{array}{r}yp\left(y\right)\left(\frac{d}{dy}p\left(y\right)\right)=0\end{array}$$

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

Factoring the ode gives these factors

$$\begin{array}{}\text{(1)}& p& =0\text{(2)}& p^{\prime }& =0\end{array}$$

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for $p$ from

$$\begin{array}{r}p=0\end{array}$$

Solving gives $p=0$

Solving equation (2)

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

$$\begin{array}{rl}\int dp& =\int 0dy+c_1\\ p& =c_1\end{array}$$

For solution (1) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=0\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int 0dx+c_2\\ y& =c_2\end{array}$$

For solution (2) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=c_1\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int c_{1}dx\\ y& =c_{1}x+c_3\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_2\\ y& =c_{1}x+c_3\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 13

ode:=y(x)*diff(diff(y(x),x),x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}y& =0\\ y& =c_{1}x+c_2\end{array}$$

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & y\left(x\right)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=0\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{0\pm \left(\sqrt{0}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=0\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(x\right)=1\\ \bullet & & \text{Repeated root, multiply}{y}_1\left(x\right)\text{by}x\text{to ensure linear independence}\\ & & y_2\left(x\right)=x\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathit{C}\mathit{2}x+\mathit{C}\mathit{1}\end{array}$$

✓ Mathematica. Time used: 0.002 (sec). Leaf size: 17

ode=y[x]*D[y[x],{x,2}]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to 0\\ y(x)\to c_{2}x+c_1\end{array}$$

✓ Sympy. Time used: 0.146 (sec). Leaf size: 7

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

$$y\left(x\right)=C_1+C_{2}x$$