2.1.6 Problem 1.1.6

Internal problem ID [13206]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, First-Order differential equations
Problem number : 1.1.6
Date solved : Sunday, January 18, 2026 at 06:41:48 PM
CAS classification : [[_homogeneous, `class A`], _dAlembert]

2.1.6.1 Solved using first_order_ode_homog_A

0.103 (sec)

Entering first order ode homog A solver

\begin{align*} y^{\prime }&=f \left (\frac {y}{x}\right ) \\ \end{align*}

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= f \left (\frac {y}{x}\right )\tag {1} \end{align*}

An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n y)= t^n f(x,y) \]

In this case, it can be seen that both \(M=f \left (\frac {y}{x}\right )\) and \(N=1\) are both homogeneous and of the same order \(n=0\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence

\[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(y=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= f \left (u \right )\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {f \left (u \left (x \right )\right )-u \left (x \right )}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {f \left (u \left (x \right )\right )-u \left (x \right )}{x} = 0 \]

Or

\[ u^{\prime }\left (x \right ) x -f \left (u \left (x \right )\right )+u \left (x \right ) = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {f \left (u \left (x \right )\right )-u \left (x \right )}{x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {f \left (u \left (x \right )\right )-u \left (x \right )}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= f \left (u \right )-u \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{f \left (u \right )-u}\,du} &= \int { \frac {1}{x} \,dx} \\ \end{align*}

\[ \int _{}^{u \left (x \right )}\frac {1}{f \left (\tau \right )-\tau }d \tau = \ln \left (x \right )+c_1 \]

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

\[ f \left (u \right )-u=0 \]

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

\begin{align*} u \left (x \right )&=\operatorname {RootOf}\left (-f \left (\textit {\_Z} \right )+\textit {\_Z} \right ) \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.

The solution \(\operatorname {RootOf}\left (-f \left (\textit {\_Z} \right )+\textit {\_Z} \right )\) will not be used

Converting \(\int _{}^{u \left (x \right )}\frac {1}{f \left (\tau \right )-\tau }d \tau = \ln \left (x \right )+c_1\) back to \(y\) gives

\begin{align*} \int _{}^{\frac {y}{x}}\frac {1}{f \left (\tau \right )-\tau }d \tau = \ln \left (x \right )+c_1 \end{align*}

Summary of solutions found

\begin{align*} \int _{}^{\frac {y}{x}}\frac {1}{f \left (\tau \right )-\tau }d \tau &= \ln \left (x \right )+c_1 \\ \end{align*}

2.1.6.2 Solved using first_order_ode_homog_type_D2

0.079 (sec)

Entering first order ode homog type D2 solver

\begin{align*} y^{\prime }&=f \left (\frac {y}{x}\right ) \\ \end{align*}

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = f \left (u \left (x \right )\right ) \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = -\frac {u \left (x \right )-f \left (u \left (x \right )\right )}{x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )-f \left (u \left (x \right )\right )}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= -u +f \left (u \right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{-u +f \left (u \right )}\,du} &= \int { \frac {1}{x} \,dx} \\ \end{align*}

\[ \int _{}^{u \left (x \right )}\frac {1}{-\tau +f \left (\tau \right )}d \tau = \ln \left (x \right )+c_1 \]

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

\[ -u +f \left (u \right )=0 \]

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

\begin{align*} u \left (x \right )&=\operatorname {RootOf}\left (-f \left (\textit {\_Z} \right )+\textit {\_Z} \right ) \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.

The solution \(\operatorname {RootOf}\left (-f \left (\textit {\_Z} \right )+\textit {\_Z} \right )\) will not be used

Converting \(\int _{}^{u \left (x \right )}\frac {1}{-\tau +f \left (\tau \right )}d \tau = \ln \left (x \right )+c_1\) back to \(y\) gives

\begin{align*} \int _{}^{\frac {y}{x}}\frac {1}{-\tau +f \left (\tau \right )}d \tau = \ln \left (x \right )+c_1 \end{align*}

Summary of solutions found

\begin{align*} \int _{}^{\frac {y}{x}}\frac {1}{-\tau +f \left (\tau \right )}d \tau &= \ln \left (x \right )+c_1 \\ \end{align*}

2.1.6.3 Solved using first_order_ode_homog_type_G

0.082 (sec)

Entering first order ode homog type G solver

\begin{align*} y^{\prime }&=f \left (\frac {y}{x}\right ) \\ \end{align*}

Multiplying the right side of the ode, which is \(f \left (\frac {y}{x}\right )\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) f \left (\frac {y}{x}\right )\\ &= \frac {x f \left (\frac {y}{x}\right )}{y}\\ &= F(x,y) \end{align*}

Since \(F \left (x , y\right )\) has \(y\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= -\frac {D\left (f \right )\left (\frac {y}{x}\right ) y -x f \left (\frac {y}{x}\right )}{y}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {D\left (f \right )\left (\frac {y}{x}\right ) y -x f \left (\frac {y}{x}\right )}{y}\\ \alpha &= \frac {f_x}{f_y} \\ &=-1 \end{align*}

Since \(\alpha \) is independent of \(x,y\) then this is Homogeneous type G.

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x}} \end{align*}

Substituting the above back into \(F(x,y)\) gives

\begin{align*} F \left (z \right ) &=\frac {f \left (z \right )}{z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(y\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{y x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1-\frac {f \left (z \right )}{z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1-\frac {f \left (z \right )}{z}\right )}d z = 0 \]

Simplifying the above gives

\begin{align*} \ln \left (x \right )-c_1 -\int _{}^{\frac {y}{x}}\frac {1}{-z +f \left (z \right )}d z &= 0 \\ \end{align*}

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 -\int _{}^{\frac {y}{x}}\frac {1}{-z +f \left (z \right )}d z &= 0 \\ \end{align*}

2.1.6.4 Solved using first_order_ode_LIE

0.511 (sec)

Entering first order ode LIE solver

\begin{align*} y^{\prime }&=f \left (\frac {y}{x}\right ) \\ \end{align*}

Writing the ode as

\begin{align*} y^{\prime }&=f \left (\frac {y}{x}\right )\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+f \left (\frac {y}{x}\right ) \left (b_{3}-a_{2}\right )-f \left (\frac {y}{x}\right )^{2} a_{3}+\frac {D\left (f \right )\left (\frac {y}{x}\right ) y \left (x a_{2}+y a_{3}+a_{1}\right )}{x^{2}}-\frac {D\left (f \right )\left (\frac {y}{x}\right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{x} = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {f \left (\frac {y}{x}\right )^{2} a_{3} x^{2}+D\left (f \right )\left (\frac {y}{x}\right ) x^{2} b_{2}-D\left (f \right )\left (\frac {y}{x}\right ) x y a_{2}+D\left (f \right )\left (\frac {y}{x}\right ) x y b_{3}-D\left (f \right )\left (\frac {y}{x}\right ) y^{2} a_{3}+x^{2} f \left (\frac {y}{x}\right ) a_{2}-x^{2} f \left (\frac {y}{x}\right ) b_{3}+D\left (f \right )\left (\frac {y}{x}\right ) x b_{1}-D\left (f \right )\left (\frac {y}{x}\right ) y a_{1}-b_{2} x^{2}}{x^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -f \left (\frac {y}{x}\right )^{2} a_{3} x^{2}-D\left (f \right )\left (\frac {y}{x}\right ) x^{2} b_{2}+D\left (f \right )\left (\frac {y}{x}\right ) x y a_{2}-D\left (f \right )\left (\frac {y}{x}\right ) x y b_{3}+D\left (f \right )\left (\frac {y}{x}\right ) y^{2} a_{3}-x^{2} f \left (\frac {y}{x}\right ) a_{2}+x^{2} f \left (\frac {y}{x}\right ) b_{3}-D\left (f \right )\left (\frac {y}{x}\right ) x b_{1}+D\left (f \right )\left (\frac {y}{x}\right ) y a_{1}+b_{2} x^{2} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \left \{x, y, f \left (\frac {y}{x}\right ), D\left (f \right )\left (\frac {y}{x}\right )\right \} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \left \{x = v_{1}, y = v_{2}, f \left (\frac {y}{x}\right ) = v_{3}, D\left (f \right )\left (\frac {y}{x}\right ) = v_{4}\right \} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -v_{3}^{2} a_{3} v_{1}^{2}-v_{1}^{2} v_{3} a_{2}+v_{4} v_{1} v_{2} a_{2}+v_{4} v_{2}^{2} a_{3}-v_{4} v_{1}^{2} b_{2}+v_{1}^{2} v_{3} b_{3}-v_{4} v_{1} v_{2} b_{3}+v_{4} v_{2} a_{1}-v_{4} v_{1} b_{1}+b_{2} v_{1}^{2} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}, v_{3}, v_{4}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} -v_{3}^{2} a_{3} v_{1}^{2}+\left (b_{3}-a_{2}\right ) v_{1}^{2} v_{3}-v_{4} v_{1}^{2} b_{2}+b_{2} v_{1}^{2}+\left (-b_{3}+a_{2}\right ) v_{1} v_{2} v_{4}-v_{4} v_{1} b_{1}+v_{4} v_{2}^{2} a_{3}+v_{4} v_{2} a_{1} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} a_{1}&=0\\ a_{3}&=0\\ b_{2}&=0\\ -a_{3}&=0\\ -b_{1}&=0\\ -b_{2}&=0\\ -b_{3}+a_{2}&=0\\ b_{3}-a_{2}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= x \\ \eta &= y \\ \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Therefore

\begin{align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {y}{x}\\ &= \frac {y}{x} \end{align*}

This is easily solved to give

\begin{align*} y = c_1 x \end{align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence

\begin{align*} R &= \frac {y}{x} \end{align*}

And \(S\) is found from

\begin{align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{x} \end{align*}

Integrating gives

\begin{align*} S &= \int { \frac {dx}{T}}\\ &= \ln \left (x \right ) \end{align*}

Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= f \left (\frac {y}{x}\right ) \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= -\frac {y}{x^{2}}\\ R_{y} &= \frac {1}{x}\\ S_{x} &= \frac {1}{x}\\ S_{y} &= 0 \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {x}{x f \left (\frac {y}{x}\right )-y}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= \frac {1}{f \left (R \right )-R} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {\frac {1}{f \left (R \right )-R}\, dR}\\ S \left (R \right ) &= \int \frac {1}{f \left (R \right )-R}d R + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \int \frac {1}{f \left (R \right )-R}d R +c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \ln \left (x \right ) = \int _{}^{\frac {y}{x}}\frac {1}{f \left (\textit {\_a} \right )-\textit {\_a}}d \textit {\_a} +c_2 \end{align*}

Summary of solutions found

\begin{align*} \ln \left (x \right ) &= \int _{}^{\frac {y}{x}}\frac {1}{f \left (\textit {\_a} \right )-\textit {\_a}}d \textit {\_a} +c_2 \\ \end{align*}

2.1.6.5 ✓ Maple. Time used: 0.003 (sec). Leaf size: 27

ode:=diff(y(x),x) = f(y(x)/x); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \operatorname {RootOf}\left (-\int _{}^{\textit {\_Z}}-\frac {1}{-f \left (\textit {\_a} \right )+\textit {\_a}}d \textit {\_a} +\ln \left (x \right )+c_1 \right ) x \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=f \left (\frac {y \left (x \right )}{x}\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=f \left (\frac {y \left (x \right )}{x}\right ) \end {array} \]

2.1.6.6 ✓ Mathematica. Time used: 0.045 (sec). Leaf size: 33

ode=D[y[x],x]==f[y[x]/x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^{\frac {y(x)}{x}}\frac {1}{K[1]-f(K[1])}dK[1]=-\log (x)+c_1,y(x)\right ] \]

2.1.6.7 ✓ Sympy. Time used: 0.544 (sec). Leaf size: 24

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

\[ y{\left (x \right )} = C_{1} e^{- \int \limits ^{\frac {x}{y{\left (x \right )}}} \frac {f{\left (\frac {1}{u_{1}} \right )}}{u_{1} f{\left (\frac {1}{u_{1}} \right )} - 1}\, du_{1}} \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('1st_homogeneous_coeff_best', '1st_homogeneous_coeff_subs_indep_div_dep', '1st_homogeneous_coeff_subs_dep_div_indep', 'lie_group', '1st_homogeneous_coeff_subs_indep_div_dep_Integral', '1st_homogeneous_coeff_subs_dep_div_indep_Integral')