2.1.52 Problem 52

Internal problem ID [10310]
Book : First order enumerated odes
Section : section 1
Problem number : 52
Date solved : Thursday, November 27, 2025 at 10:32:55 AM
CAS classification : [_separable]

Solved using first_order_nonlinear_p_but_separable

Time used: 0.604 (sec)

Solve

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

The ode has the form

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

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

\begin{align*} (y')^{2} &= \frac {y^{2}}{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^{2} &> 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^{2}}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx\\ -\frac {1}{\sqrt {y^{2}}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx \end{align*}

Integrating now gives the following solutions

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

Therefore

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

Solving for \(y\) gives

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

Summary of solutions found

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

Solved using first_order_ode_parametric method

Time used: 0.572 (sec)

Solve

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

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

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

Isolating \(x\) gives

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

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

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

Which is now solved for \(y\).

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(\lambda ,y) \mathop {\mathrm {d}\lambda }+ N(\lambda ,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]

Therefore

\begin{align*} \left (\lambda \left (-\lambda +2 y \right )\right )\mathop {\mathrm {d}y} &= \left (2 y^{2}\right )\mathop {\mathrm {d}\lambda }\\ \left (-2 y^{2}\right )\mathop {\mathrm {d}\lambda } + \left (\lambda \left (-\lambda +2 y \right )\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(\lambda ,y) &= -2 y^{2}\\ N(\lambda ,y) &= \lambda \left (-\lambda +2 y \right ) \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial \lambda } \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-2 y^{2}\right )\\ &= -4 y \end{align*}

And

\begin{align*} \frac {\partial N}{\partial \lambda } &= \frac {\partial }{\partial \lambda } \left (\lambda \left (-\lambda +2 y \right )\right )\\ &= -2 \lambda +2 y \end{align*}

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial \lambda }\), then the ODE is not exact. By inspection \(\frac {1}{\lambda ^{2} y \left (\lambda \right )}\) is an integrating factor. Therefore by multiplying \(M=-2 y \left (\lambda \right )^{2}\) and \(N=\lambda \left (-\lambda +2 y \left (\lambda \right )\right )\) by this integrating factor the ode becomes exact. The new \(M,N\) are

\begin{align*} M&=-\frac {2 y \left (\lambda \right )}{\lambda ^{2}} \\ N&=\frac {-\lambda +2 y \left (\lambda \right )}{\lambda y \left (\lambda \right )} \\ \end{align*}

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(\lambda ,y) \mathop {\mathrm {d}\lambda }+ N(\lambda ,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]

Therefore

\begin{align*} \left (\frac {-\lambda +2 y}{\lambda y}\right )\mathop {\mathrm {d}y} &= \left (\frac {2 y}{\lambda ^{2}}\right )\mathop {\mathrm {d}\lambda }\\ \left (-\frac {2 y}{\lambda ^{2}}\right )\mathop {\mathrm {d}\lambda } + \left (\frac {-\lambda +2 y}{\lambda y}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(\lambda ,y) &= -\frac {2 y}{\lambda ^{2}}\\ N(\lambda ,y) &= \frac {-\lambda +2 y}{\lambda y} \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial \lambda } \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\frac {2 y}{\lambda ^{2}}\right )\\ &= -\frac {2}{\lambda ^{2}} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial \lambda } &= \frac {\partial }{\partial \lambda } \left (\frac {-\lambda +2 y}{\lambda y}\right )\\ &= -\frac {2}{\lambda ^{2}} \end{align*}

Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial \lambda }\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (\lambda ,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial \lambda } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end{align*}

Integrating (1) w.r.t. \(\lambda \) gives

\begin{align*} \int \frac {\partial \phi }{\partial \lambda } \mathop {\mathrm {d}\lambda } &= \int M\mathop {\mathrm {d}\lambda } \\ \int \frac {\partial \phi }{\partial \lambda } \mathop {\mathrm {d}\lambda } &= \int -\frac {2 y}{\lambda ^{2}}\mathop {\mathrm {d}\lambda } \\ \tag{3} \phi &= \frac {2 y}{\lambda }+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(\lambda \) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = \frac {2}{\lambda }+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {-\lambda +2 y}{\lambda y}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {-\lambda +2 y}{\lambda y} = \frac {2}{\lambda }+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = -\frac {1}{y} \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {1}{y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\ln \left (y \right )+ c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {2 y}{\lambda }-\ln \left (y \right )+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = \frac {2 y}{\lambda }-\ln \left (y \right ) \]

Solving for \(y \left (\lambda \right )\) gives

\begin{align*} y \left (\lambda \right ) &= {\mathrm e}^{-\operatorname {LambertW}\left (-\frac {2 \,{\mathrm e}^{-c_1}}{\lambda }\right )-c_1} \\ \end{align*}

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

\begin{align*} y &= {\mathrm e}^{-\operatorname {LambertW}\left (-\frac {2 \,{\mathrm e}^{-c_1}}{\lambda }\right )-c_1} \\ x &= \frac {y^{2}}{\lambda ^{2}} \\ \end{align*}

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

\[ -y+{\mathrm e}^{-\operatorname {LambertW}\left (-\frac {2 \operatorname {RootOf}\left (x \,\textit {\_Z}^{2}-1\right ) x \,{\mathrm e}^{-c_1}}{y}\right )-c_1} \]

Which can be written as

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

Solving for \(y\) gives

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

Summary of solutions found

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

✓ Maple. Time used: 0.021 (sec). Leaf size: 27

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

\begin{align*} y &= 0 \\ y &= c_1 \,{\mathrm e}^{-2 \sqrt {x}} \\ y &= c_1 \,{\mathrm e}^{2 \sqrt {x}} \\ \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 simple symmetries for implicit equations 
<- symmetries for implicit equations 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 )^{2}}{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 {y \left (x \right )}{\sqrt {x}}, \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{\sqrt {x}}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right )}{\sqrt {x}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}=\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}d x =\int \frac {1}{\sqrt {x}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (x \right )\right )=2 \sqrt {x}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{2 \sqrt {x}+\textit {\_C1}} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\textit {\_C1} \,{\mathrm e}^{2 \sqrt {x}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{\sqrt {x}} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}=-\frac {1}{\sqrt {x}} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )}d x =\int -\frac {1}{\sqrt {x}}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (x \right )\right )=-2 \sqrt {x}+\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )={\mathrm e}^{-2 \sqrt {x}+\textit {\_C1}} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\textit {\_C1} \,{\mathrm e}^{-2 \sqrt {x}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\mathit {C1} \,{\mathrm e}^{-2 \sqrt {x}}, y \left (x \right )=\mathit {C1} \,{\mathrm e}^{2 \sqrt {x}}\right \} \end {array} \]

✓ Mathematica. Time used: 0.042 (sec). Leaf size: 38

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

\begin{align*} y(x)&\to c_1 e^{-2 \sqrt {x}}\\ y(x)&\to c_1 e^{2 \sqrt {x}}\\ y(x)&\to 0 \end{align*}

✓ Sympy. Time used: 0.266 (sec). Leaf size: 31

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

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