2.1.5 Problem 5

Internal problem ID [9076]
Book : Second order enumerated odes
Section : section 1
Problem number : 5
Date solved : Sunday, March 30, 2025 at 02:06:19 PM
CAS classification : [[_2nd_order, _quadrature]]

Solved as second order missing x ode

Time used: 0.261 (sec)

Solve

$$\begin{array}{rl}a{y^{\prime \prime }}^2& =0\end{array}$$

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

$$\begin{array}{r}ap{\left(y\right)}^2{\left(\frac{d}{dy}p\left(y\right)\right)}^2=0\end{array}$$

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

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

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

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

Solving Eq. (2)

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

$$\begin{array}{rl}\int dp& =\int 0dy+c_2\\ p& =c_2\end{array}$$

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

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

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

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

Solved as second order missing y ode

Time used: 0.216 (sec)

Solve

$$\begin{array}{rl}a{y^{\prime \prime }}^2& =0\end{array}$$

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

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

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

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

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

$$\begin{array}{rl}\int du& =\int 0dx+c_1\\ u& =c_1\end{array}$$

Solving Eq. (2)

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

$$\begin{array}{rl}\int du& =\int 0dx+c_2\\ u& =c_2\end{array}$$

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

$$\begin{array}{rl}u& =c_1\end{array}$$

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

$$\begin{array}{r}{y}^{\prime }\left(x\right)=c_1\end{array}$$

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

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

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

✓ Maple. Time used: 0.011 (sec). Leaf size: 9

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

$$y=c_{1}x+c_2$$

Maple trace

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

Maple step by step

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

✓ Mathematica. Time used: 0.002 (sec). Leaf size: 12

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

$$y(x)\to c_{2}x+c_1$$

✓ Sympy. Time used: 0.039 (sec). Leaf size: 7

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

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