2.1.79 Problem 78

Internal problem ID [10065]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 78
Date solved : Monday, December 29, 2025 at 09:04:36 PM
CAS classification : [[_homogeneous, `class A`], _dAlembert]

2.1.79.1 Solved using first_order_ode_dAlembert

1.503 (sec)

Entering first order ode dAlembert solver

\begin{align*} \frac {y^{\prime } y}{1+\frac {\sqrt {1+{y^{\prime }}^{2}}}{2}}&=-x \\ y \left (0\right ) &= 3 \\ \end{align*}

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

\begin{align*} \frac {p y}{1+\frac {\sqrt {p^{2}+1}}{2}} = -x \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= -\frac {x \left (2+\sqrt {p^{2}+1}\right )}{2 p} \\ \end{align*}

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

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

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {-2-\sqrt {p^{2}+1}}{2 p}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {-2-\sqrt {p^{2}+1}}{2 p} = \left (-\frac {x}{2 \sqrt {p^{2}+1}}+\frac {x}{p^{2}}+\frac {x \sqrt {p^{2}+1}}{2 p^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

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

\begin{align*} p -\frac {-2-\sqrt {p^{2}+1}}{2 p} = 0 \end{align*}

No valid singular solutions found.

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

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {-2-\sqrt {p \left (x \right )^{2}+1}}{2 p \left (x \right )}}{-\frac {x}{2 \sqrt {p \left (x \right )^{2}+1}}+\frac {x}{p \left (x \right )^{2}}+\frac {x \sqrt {p \left (x \right )^{2}+1}}{2 p \left (x \right )^{2}}} \end{equation}

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

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {\left (2 p \left (x \right )^{2}+\sqrt {p \left (x \right )^{2}+1}+2\right ) \sqrt {p \left (x \right )^{2}+1}\, p \left (x \right )}{x \left (1+2 \sqrt {p \left (x \right )^{2}+1}\right )} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\left (2 p \left (x \right )^{2}+\sqrt {p \left (x \right )^{2}+1}+2\right ) \sqrt {p \left (x \right )^{2}+1}\, p \left (x \right )}{x \left (1+2 \sqrt {p \left (x \right )^{2}+1}\right )}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(p) &= \frac {\left (2 p^{2}+\sqrt {p^{2}+1}+2\right ) \sqrt {p^{2}+1}\, p}{1+2 \sqrt {p^{2}+1}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {1+2 \sqrt {p^{2}+1}}{\left (2 p^{2}+\sqrt {p^{2}+1}+2\right ) \sqrt {p^{2}+1}\, p}\,dp} &= \int { \frac {1}{x} \,dx} \\ \end{align*}

\[ \ln \left (\frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}}\right )=\ln \left (x \right )+c_1 \]

Taking the exponential of both sides the solution becomes

\[ \frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}} = c_1 x \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (2 p^{2}+\sqrt {p^{2}+1}+2\right ) \sqrt {p^{2}+1}\, p}{1+2 \sqrt {p^{2}+1}}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=0\\ p \left (x \right )&=-i \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}} &= c_1 x \\ p \left (x \right ) &= 0 \\ p \left (x \right ) &= -i \\ \end{align*}

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

\begin{align*} y &= \frac {-2-\sqrt {-\frac {c_1^{2} x^{2}}{c_1^{2} x^{2}-1}+1}}{2 c_1 \sqrt {-\frac {1}{c_1^{2} x^{2}-1}}} \\ y &= -i x \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= -\frac {2+\sqrt {-\frac {1}{c_1^{2} x^{2}-1}}}{2 \sqrt {-\frac {1}{c_1^{2} x^{2}-1}}\, c_1} \\ y &= -i x \\ \end{align*}

Solving for initial conditions the solution is

\begin{align*} y &= \frac {1+\sqrt {-\frac {1}{x^{2}-4}}}{\sqrt {-\frac {1}{x^{2}-4}}} \\ y &= -i x \\ \end{align*}

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

Figure 2.161: Solutions plot

Summary of solutions found

\begin{align*} y &= \frac {1+\sqrt {-\frac {1}{x^{2}-4}}}{\sqrt {-\frac {1}{x^{2}-4}}} \\ \end{align*}

2.1.79.2 ✓ Maple. Time used: 4.414 (sec). Leaf size: 29

ode:=diff(y(x),x)*y(x)/(1+1/2*(1+diff(y(x),x)^2)^(1/2)) = -x; 
ic:=[y(0) = 3]; 
dsolve([ode,op(ic)],y(x), singsol=all);
 

\begin{align*} y &= -3+\sqrt {-x^{2}+36} \\ y &= 1+\sqrt {-x^{2}+4} \\ \end{align*}

Maple trace

Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying dAlembert 
<- dAlembert successful
 

2.1.79.3 ✓ Mathematica. Time used: 0.366 (sec). Leaf size: 35

ode=D[y[x],x]*y[x]/(1+1/2*Sqrt[1+(D[y[x],x])^2])==-x; 
ic=y[0]==3; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \sqrt {4-x^2}+1\\ y(x)&\to \sqrt {36-x^2}-3 \end{align*}

2.1.79.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x + y(x)*Derivative(y(x), x)/(sqrt(Derivative(y(x), x)**2 + 1)/2 + 1),0) 
ics = {y(0): 3} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out