2.1.8 Problem 8

Internal problem ID [9079]
Book : Second order enumerated odes
Section : section 1
Problem number : 8
Date solved : Sunday, March 30, 2025 at 02:06:24 PM
CAS classification : [[_2nd_order, _quadrature]]

Solved as second order missing x ode

Time used: 0.665 (sec)

Solve

$$\begin{array}{rl}{y^{\prime \prime }}^2& =1\end{array}$$

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

$$\begin{array}{r}p{\left(y\right)}^2{\left(\frac{d}{dy}p\left(y\right)\right)}^2=1\end{array}$$

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

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

$$\begin{array}{rl}\int pdp& =dy\\ \frac{p^2}{2}& =y+c_1\end{array}$$

Solving Eq. (2)

Integrating gives

$$\begin{array}{rl}\int -pdp& =dy\\ -\frac{p^2}{2}& =y+c_2\end{array}$$

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

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

Let $p=y^{\prime }$ the ode becomes

$$\begin{array}{r}\frac{p^2}{2}=y+c_1\end{array}$$

Solving for $y$ from the above results in

$$\begin{array}{}\text{(1)}& y& =-c_1+\frac{p^2}{2}\end{array}$$

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =-c_1+\frac{p^2}{2}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=p{p}^{\prime }\left(x\right)\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

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

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=-c_1\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=1\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

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

$$\begin{array}{rl}\int dp& =\int 1dx\\ p\left(x\right)& =x+c_3\end{array}$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{rl}y& =-c_1+\frac{1}{2}{c}_3^2+c_{3}x+\frac{1}{2}{x}^2\end{array}$$

Will add steps showing solving for IC soon.

The solution

$$y=-c_1$$

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

$$\begin{array}{rl}y& =-c_1+\frac{1}{2}{c}_3^2+c_{3}x+\frac{1}{2}{x}^2\end{array}$$

Solved as second order missing y ode

Time used: 0.252 (sec)

Solve

$$\begin{array}{rl}{y^{\prime \prime }}^2& =1\end{array}$$

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

$$\begin{array}{r}{u^{\prime }\left(x\right)}^2-1=0\end{array}$$

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

Solving for the derivative gives these ODE’s to solve

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

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 $f(x)$.

$$\begin{array}{rl}\int du& =\int 1dx\\ u& =x+c_1\end{array}$$

Solving Eq. (2)

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

$$\begin{array}{rl}\int du& =\int -1dx\\ u& =-x+c_2\end{array}$$

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

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

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

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

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

$$\begin{array}{rl}\int dy& =\int x+c_{1}dx\\ y& =\frac{1}{2}{x}^2+c_{1}x+c_3\end{array}$$

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

✓ Maple. Time used: 0.002 (sec). Leaf size: 27

ode:=diff(diff(y(x),x),x)^2 = 1; 
dsolve(ode,y(x), singsol=all);
 

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

Maple trace

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & {\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)}^2=1\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Make substitution}u=\frac{d}{dx}y\left(x\right)\text{to reduce order of ODE}\\ & & {\left(\frac{d}{dx}u\left(x\right)\right)}^2=1\\ \bullet & & \text{Solve for the highest derivative}\\ & & [\frac{d}{dx}u\left(x\right)=-1,\frac{d}{dx}u\left(x\right)=1]\\ ◻& & \text{Solve the equation}\frac{d}{dx}u\left(x\right)=-1\\ & \circ & \text{Integrate both sides with respect to}x\\ & & \int \left(\frac{d}{dx}u\left(x\right)\right)dx=\int (-1)dx+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & u\left(x\right)=-x+\mathit{\text{\_C1}}\\ ◻& & \text{Solve the equation}\frac{d}{dx}u\left(x\right)=1\\ & \circ & \text{Integrate both sides with respect to}x\\ & & \int \left(\frac{d}{dx}u\left(x\right)\right)dx=\int 1dx+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & u\left(x\right)=x+\mathit{\text{\_C1}}\\ \bullet & & \text{Set of solutions}\\ & & \{u\left(x\right)=-x+\mathit{C}\mathit{1},u\left(x\right)=x+\mathit{C}\mathit{1}\}\\ \bullet & & \text{Define new dependent variable}u\\ & & u\left(x\right)=\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Compute}\frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ & & \frac{d}{dx}u\left(x\right)=\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Use chain rule on the lhs}\\ & & \left(\frac{d}{dx}y\left(x\right)\right)\left(\frac{d}{dy}u\left(y\right)\right)=\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Substitute in the definition of}u\\ & & u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)=\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Make substitutions}\frac{d}{dx}y\left(x\right)=u\left(y\right),\frac{d}{dx}\frac{d}{dx}y\left(x\right)=u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)\text{to reduce order of ODE}\\ & & u{\left(y\right)}^2{\left(\frac{d}{dy}u\left(y\right)\right)}^2=1\\ \bullet & & \text{Solve for the highest derivative}\\ & & [\frac{d}{dy}u\left(y\right)=\frac{1}{u\left(y\right)},\frac{d}{dy}u\left(y\right)=-\frac{1}{u\left(y\right)}]\\ ◻& & \text{Solve the equation}\frac{d}{dy}u\left(y\right)=\frac{1}{u\left(y\right)}\\ & \circ & \text{Separate variables}\\ & & u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)=1\\ & \circ & \text{Integrate both sides with respect to}y\\ & & \int u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)dy=\int 1dy+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & \frac{u{\left(y\right)}^2}{2}=y+\mathit{\text{\_C1}}\\ & \circ & \text{Solve for}u\left(y\right)\\ & & \{u\left(y\right)=\sqrt{2\mathit{\text{\_C1}}+2y},u\left(y\right)=-\sqrt{2\mathit{\text{\_C1}}+2y}\}\\ & \circ & \text{Redefine the integration constant(s)}\\ & & \{u\left(y\right)=\sqrt{\mathit{\text{\_C1}}+2y},u\left(y\right)=-\sqrt{\mathit{\text{\_C1}}+2y}\}\\ ◻& & \text{Solve the equation}\frac{d}{dy}u\left(y\right)=-\frac{1}{u\left(y\right)}\\ & \circ & \text{Separate variables}\\ & & u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)=-1\\ & \circ & \text{Integrate both sides with respect to}y\\ & & \int u\left(y\right)\left(\frac{d}{dy}u\left(y\right)\right)dy=\int (-1)dy+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & \frac{u{\left(y\right)}^2}{2}=-y+\mathit{\text{\_C1}}\\ & \circ & \text{Solve for}u\left(y\right)\\ & & \{u\left(y\right)=\sqrt{2\mathit{\text{\_C1}}-2y},u\left(y\right)=-\sqrt{2\mathit{\text{\_C1}}-2y}\}\\ & \circ & \text{Redefine the integration constant(s)}\\ & & \{u\left(y\right)=\sqrt{\mathit{\text{\_C1}}-2y},u\left(y\right)=-\sqrt{\mathit{\text{\_C1}}-2y}\}\\ \bullet & & \text{Set of solutions}\\ & & \{u\left(y\right)=\sqrt{\mathit{C}\mathit{1}-2y},u\left(y\right)=\sqrt{\mathit{C}\mathit{1}+2y},u\left(y\right)=-\sqrt{\mathit{C}\mathit{1}-2y},u\left(y\right)=-\sqrt{\mathit{C}\mathit{1}+2y}\}\\ \bullet & & \text{Solve 1st ODE for}u\left(y\right)\\ & & u\left(y\right)=\sqrt{\mathit{C}\mathit{1}-2y}\\ \bullet & & \text{Revert to original variables with substitution}u\left(y\right)=\frac{d}{dx}y\left(x\right),y=y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}}=1\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}}dx=\int 1dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Evaluate integral}\\ & & -\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}=x+\mathit{C}\mathit{2}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=-\frac{1}{2}{\mathit{C}\mathit{2}}^2-\mathit{C}\mathit{2}x-\frac{1}{2}{x}^2+\frac{1}{2}\mathit{C}\mathit{1}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\mathit{C}\mathit{2}x-\frac{1}{2}{x}^2+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve 2nd ODE for}u\left(y\right)\\ & & u\left(y\right)=\sqrt{\mathit{C}\mathit{1}+2y}\\ \bullet & & \text{Revert to original variables with substitution}u\left(y\right)=\frac{d}{dx}y\left(x\right),y=y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}}=1\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}}dx=\int 1dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Evaluate integral}\\ & & \sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}=x+\mathit{C}\mathit{2}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=\frac{1}{2}{\mathit{C}\mathit{2}}^2+\mathit{C}\mathit{2}x+\frac{1}{2}{x}^2-\frac{1}{2}\mathit{C}\mathit{1}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\mathit{C}\mathit{2}x+\frac{1}{2}{x}^2+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve 3rd ODE for}u\left(y\right)\\ & & u\left(y\right)=-\sqrt{\mathit{C}\mathit{1}-2y}\\ \bullet & & \text{Revert to original variables with substitution}u\left(y\right)=\frac{d}{dx}y\left(x\right),y=y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=-\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=-\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}}=-1\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}}dx=\int (-1)dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Evaluate integral}\\ & & -\sqrt{\mathit{C}\mathit{1}-2y\left(x\right)}=-x+\mathit{C}\mathit{2}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=-\frac{1}{2}{\mathit{C}\mathit{2}}^2+\mathit{C}\mathit{2}x-\frac{1}{2}{x}^2+\frac{1}{2}\mathit{C}\mathit{1}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\mathit{C}\mathit{2}x-\frac{1}{2}{x}^2+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve 4th ODE for}u\left(y\right)\\ & & u\left(y\right)=-\sqrt{\mathit{C}\mathit{1}+2y}\\ \bullet & & \text{Revert to original variables with substitution}u\left(y\right)=\frac{d}{dx}y\left(x\right),y=y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=-\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=-\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}}=-1\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{\sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}}dx=\int (-1)dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Evaluate integral}\\ & & \sqrt{\mathit{C}\mathit{1}+2y\left(x\right)}=-x+\mathit{C}\mathit{2}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)=\frac{1}{2}{\mathit{C}\mathit{2}}^2-\mathit{C}\mathit{2}x+\frac{1}{2}{x}^2-\frac{1}{2}\mathit{C}\mathit{1}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\mathit{C}\mathit{2}x+\frac{1}{2}{x}^2+\mathit{C}\mathit{1}\end{array}$$

✓ Mathematica. Time used: 0.003 (sec). Leaf size: 37

ode=(D[y[x],{x,2}])^2==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

✓ Sympy. Time used: 0.151 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), (x, 2))**2 - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

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