2.1.17 Problem 17

Internal problem ID [9088]
Book : Second order enumerated odes
Section : section 1
Problem number : 17
Date solved : Sunday, March 30, 2025 at 02:06:41 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]

Solved as second order missing x ode

Time used: 1.394 (sec)

Solve

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

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

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

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

Integrating gives

$$\begin{array}{rl}\int -\frac{p}{p^2-1}dp& =dy\\ -\frac{\ln (p^2-1)}{2}& =y+c_1\end{array}$$

Applying the exponential to both sides gives

$$\begin{array}{rl}{e}^{\ln \left(\frac{1}{\sqrt{p^2-1}}\right)}& =e^{y+c_1}\\ \frac{1}{\sqrt{p^2-1}}& =c_1{\mathrm{e}}^y\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-\frac{p^2-1}{p}& =0\end{array}$$

for $p$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}p=-1\\ p=1\end{array}$$

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

$$\begin{array}{r}\frac{1}{\sqrt{{y^{\prime }}^2-1}}=c_1{\mathrm{e}}^y\end{array}$$

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

$$\begin{array}{r}\frac{1}{\sqrt{p^2-1}}=c_1{\mathrm{e}}^y\end{array}$$

Solving for $y$ from the above results in

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

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

$$\begin{array}{rl}f& =0\\ g& =-\frac{\ln \left(c_1^2(p^2-1)\right)}{2}\end{array}$$

Hence (2) becomes

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

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

$$\begin{array}{r}y=-\frac{\ln (-c_1^2)}{2}\end{array}$$

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

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=-p{\left(x\right)}^2+1\end{array}$$

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

Integrating gives

$$\begin{array}{rl}\int \frac{1}{-p^2+1}dp& =dx\\ \mathrm{arctanh}\left(p\right)& =x+c_2\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-p^2+1& =0\end{array}$$

for $p\left(x\right)$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}p\left(x\right)=-1\\ p\left(x\right)=1\end{array}$$

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

$$\begin{array}{rl}y& =-\frac{\ln \left(c_1^2(\tanh {(x+c_2)}^2-1)\right)}{2}\end{array}$$

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

$$\begin{array}{r}{y}^{\prime }=-1\end{array}$$

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

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

For solution (3) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=1\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int 1dx\\ y& =x+c_4\end{array}$$

Will add steps showing solving for IC soon.

The solution

$$y=-\frac{\ln (-c_1^2)}{2}$$

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

Summary of solutions found

$$\begin{array}{rl}y& =-\frac{\ln \left(c_1^2(\tanh {(x+c_2)}^2-1)\right)}{2}\\ y& =-x+c_3\\ y& =x+c_4\end{array}$$

Solved as second order missing y ode

Time used: 0.218 (sec)

Solve

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

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

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

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

Integrating gives

$$\begin{array}{rl}\int \frac{1}{-u^2+1}du& =dx\\ \mathrm{arctanh}\left(u\right)& =x+c_1\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-u^2+1& =0\end{array}$$

for $u$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}u=-1\\ u=1\end{array}$$

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

$$\begin{array}{rl}\mathrm{arctanh}\left(u\right)& =x+c_1\\ u& =-1\\ u& =1\end{array}$$

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

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

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

$$\begin{array}{rl}\int dy& =\int \tanh (x+c_1)dx\\ y& =\ln (\cosh (x+c_1))+c_2\end{array}$$

For solution $u\left(x\right)=-1$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=-1\end{array}$$

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

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

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

$$\begin{array}{r}{y}^{\prime }=1\end{array}$$

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

$$\begin{array}{rl}\int dy& =\int 1dx\\ y& =x+c_4\end{array}$$

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

$$\begin{array}{rl}y& =\ln (\cosh (x+c_1))+c_2\\ y& =-x+c_3\\ y& =x+c_4\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =-x+c_3\\ y& =x+c_4\\ y& =\ln (\cosh (x+c_1))+c_2\end{array}$$

✓ Maple. Time used: 0.018 (sec). Leaf size: 22

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

$$y=-x-\ln \left(2\right)+\ln (-c_1{\mathrm{e}}^{2x}+c_2)$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+{\left(\frac{d}{dx}y\left(x\right)\right)}^2=1\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Make substitution}u=\frac{d}{dx}y\left(x\right)\text{to reduce order of ODE}\\ & & \frac{d}{dx}u\left(x\right)+u{\left(x\right)}^2=1\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}u\left(x\right)=-u{\left(x\right)}^2+1\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}u\left(x\right)}{-u{\left(x\right)}^2+1}=1\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}u\left(x\right)}{-u{\left(x\right)}^2+1}dx=\int 1dx+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \mathrm{arctanh}\left(u\left(x\right)\right)=x+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}u\left(x\right)\\ & & u\left(x\right)=\tanh (x+\mathit{C}\mathit{1})\\ \bullet & & \text{Solve 1st ODE for}u\left(x\right)\\ & & u\left(x\right)=\tanh (x+\mathit{C}\mathit{1})\\ \bullet & & \text{Make substitution}u=\frac{d}{dx}y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=\tanh (x+\mathit{C}\mathit{1})\\ \bullet & & \text{Integrate both sides to solve for}y\left(x\right)\\ & & \int \left(\frac{d}{dx}y\left(x\right)\right)dx=\int \tanh (x+\mathit{C}\mathit{1})dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Compute integrals}\\ & & y\left(x\right)=\ln (\cosh (x+\mathit{C}\mathit{1}))+\mathit{C}\mathit{2}\end{array}$$

✓ Mathematica. Time used: 0.354 (sec). Leaf size: 48

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

$$\begin{array}{r}y(x)\to -\frac{1}{2}\log \left(e^{2x}\right)+\log (e^{2x}+e^{2c_1})+c_2\\ y(x)\to \frac{1}{2}\log \left(e^{2x}\right)+c_2\end{array}$$

✗ Sympy

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

Timed Out