2.1.51 Problem 51

Internal problem ID [10309]
Book : First order enumerated odes
Section : section 1
Problem number : 51
Date solved : Thursday, November 27, 2025 at 10:32:52 AM
CAS classification : [[_homogeneous, `class A`], _rational, _dAlembert]

Solved using first_order_ode_dAlembert

Time used: 0.139 (sec)

Solve

\begin{align*} {y^{\prime }}^{2}&=\frac {y}{x} \\ \end{align*}

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

\begin{align*} p^{2} = \frac {y}{x} \end{align*}

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

\begin{align*} \tag{1} y &= p^{2} x \\ \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 &= p^{2}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} -p^{2}+p = 2 x p 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^{2}+p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0\\ p_{2} &=1 \end{align*}

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

\begin{align*} y = 0\\ y = x \end{align*}

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 )^{2}+p \left (x \right )}{2 x p \left (x \right )} \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 {p \left (x \right )-1}{2 x} \end{equation}

is separable as it can be written as

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

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(p) &= -\frac {p}{2}+\frac {1}{2} \end{align*}

Integrating gives

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

\[ -2 \ln \left (p \left (x \right )-1\right )=\ln \left (x \right )+c_1 \]

Taking the exponential of both sides the solution becomes

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

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

\[ -\frac {p}{2}+\frac {1}{2}=0 \]

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

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

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

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

Summary of solutions found

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

Solved using first_order_ode_homog_type_G

Time used: 0.173 (sec)

Solve

\begin{align*} {y^{\prime }}^{2}&=\frac {y}{x} \\ \end{align*}

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

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \frac {\sqrt {x y}}{x}\\ &= \frac {\sqrt {x y}}{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 {x}{2 \sqrt {x y}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {x}{2 \sqrt {x 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 {1}{\sqrt {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

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

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

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) -\frac {\sqrt {x y}}{x}\\ &= -\frac {\sqrt {x y}}{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 {x}{2 \sqrt {x y}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {x}{2 \sqrt {x 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 {1}{\sqrt {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

\[ \ln \left (x \right )-c_2 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1+\frac {1}{\sqrt {z}}\right )}d z = 0 \]

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1-\frac {1}{\sqrt {z}}\right )}d z &= 0 \\ \ln \left (x \right )-c_2 +\int _{}^{\frac {y}{x}}\frac {1}{z \left (1+\frac {1}{\sqrt {z}}\right )}d z &= 0 \\ \end{align*}

Solved using first_order_nonlinear_p_but_separable

Time used: 0.382 (sec)

Solve

\begin{align*} {y^{\prime }}^{2}&=\frac {y}{x} \\ \end{align*}

The ode has the form

\begin{align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end{align*}

Where \(n=2, m=1, f=\frac {1}{x} , g=y\). Hence the ode is

\begin{align*} (y')^{2} &= \frac {y}{x} \end{align*}

Solving for \(y^{\prime }\) from (1) gives

\begin{align*} y^{\prime } &=\sqrt {f g}\\ y^{\prime } &=-\sqrt {f g} \end{align*}

To be able to solve as separable ode, we have to now assume that \(f>0,g>0\).

\begin{align*} \frac {1}{x} &> 0\\ y &> 0 \end{align*}

Under the above assumption the differential equations become separable and can be written as

\begin{align*} y^{\prime } &=\sqrt {f}\, \sqrt {g}\\ y^{\prime } &=-\sqrt {f}\, \sqrt {g} \end{align*}

Therefore

\begin{align*} \frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx\\ -\frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx \end{align*}

Replacing \(f(x),g(y)\) by their values gives

\begin{align*} \frac {1}{\sqrt {y}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx\\ -\frac {1}{\sqrt {y}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx \end{align*}

Integrating now gives the following solutions

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

Therefore

\begin{align*} y &= x \sqrt {\frac {1}{x}}\, c_1 +\frac {c_1^{2}}{4}+x \\ y &= x \sqrt {\frac {1}{x}}\, c_1 +\frac {c_1^{2}}{4}+x \\ \end{align*}

Summary of solutions found

\begin{align*} y &= x \sqrt {\frac {1}{x}}\, c_1 +\frac {c_1^{2}}{4}+x \\ \end{align*}

Solved using first_order_ode_parametric method

Time used: 0.099 (sec)

Solve

\begin{align*} {y^{\prime }}^{2}&=\frac {y}{x} \\ \end{align*}

Let \(y^{\prime }\) be a parameter \(\lambda \). The ode becomes

\begin{align*} \lambda ^{2}-\frac {y}{x} = 0 \end{align*}

Isolating \(y\) gives

\begin{align*} y&=\lambda ^{2} x\\ &=\lambda ^{2} x\\ &=F \left (x , \lambda \right ) \end{align*}

Now we generate an ode in \(x \left (\lambda \right )\) using

\begin{align*} \frac {d}{d \lambda }x \left (\lambda \right ) &= \frac { \frac {\partial F}{\partial \lambda }} { \lambda -\frac {\partial F}{\partial x} } \\ &= \frac {2 \lambda x}{-\lambda ^{2}+\lambda }\\ &= -\frac {2 x \left (\lambda \right )}{\lambda -1} \end{align*}

Which is now solved for \(x\).

Solve In canonical form a linear first order is

\begin{align*} \frac {d}{d \lambda }x \left (\lambda \right ) + q(\lambda )x \left (\lambda \right ) &= p(\lambda ) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(\lambda ) &=\frac {2}{\lambda -1}\\ p(\lambda ) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,d\lambda }}\\ &= {\mathrm e}^{\int \frac {2}{\lambda -1}d \lambda }\\ &= \left (\lambda -1\right )^{2} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\lambda }} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\lambda }} \left (x \left (\lambda -1\right )^{2}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} x \left (\lambda -1\right )^{2}&= \int {0 \,d\lambda } + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\left (\lambda -1\right )^{2}\) gives the final solution

\[ x \left (\lambda \right ) = \frac {c_1}{\left (\lambda -1\right )^{2}} \]

Now that we found solution \(x\) we have two equations with parameter \(\lambda \). They are

\begin{align*} y &= \lambda ^{2} x \\ x &= \frac {c_1}{\left (\lambda -1\right )^{2}} \\ \end{align*}

Eliminating \(\lambda \) gives the solution for \(y\).

Summary of solutions found

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

✓ Maple. Time used: 0.017 (sec). Leaf size: 39

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

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

Maple trace

Methods for first order ODEs: 
   *** Sublevel 2 *** 
   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
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}=\frac {y \left (x \right )}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\sqrt {x y \left (x \right )}}{x}, \frac {d}{d x}y \left (x \right )=-\frac {\sqrt {x y \left (x \right )}}{x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {\sqrt {x y \left (x \right )}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\sqrt {x y \left (x \right )}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

✓ Mathematica. Time used: 0.028 (sec). Leaf size: 46

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

\begin{align*} y(x)&\to \frac {1}{4} \left (-2 \sqrt {x}+c_1\right ){}^2\\ y(x)&\to \frac {1}{4} \left (2 \sqrt {x}+c_1\right ){}^2\\ y(x)&\to 0 \end{align*}

✓ Sympy. Time used: 0.426 (sec). Leaf size: 15

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

\[ y{\left (x \right )} = \frac {C_{1}^{2}}{4} - C_{1} \sqrt {x} + x \]