2.1.52 Problem 52

Internal problem ID [9035]
Book : First order enumerated odes
Section : section 1
Problem number : 52
Date solved : Friday, April 25, 2025 at 05:34:57 PM
CAS classification : [_separable]

Solved using first_order_ode_parametric method

Time used: 1.124 (sec)

Solve

$$\begin{array}{rl}{y^{\prime }}^2& =\frac{y^2}{x}\end{array}$$

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

$$\begin{array}{r}{\lambda }^2-\frac{y^2}{x}=0\end{array}$$

Isolating $x$ gives

$$\begin{array}{r}x=\frac{y^2}{{\lambda }^2}\\ x=F(y,\lambda )\end{array}$$

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

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

Which is now solved for $y$.

In canonical form, the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(\lambda ,y)\\ \text{(1)}& & =\frac{2y^2}{\lambda (-\lambda +2y)}\end{array}$$

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

$$f(t^n\lambda ,t^{n}y)=t^{n}f(\lambda ,y)$$

In this case, it can be seen that both $M=-2y^2$ and $N=\lambda (\lambda -2y)$ are both homogeneous and of the same order $n=2$. Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution $u=\frac{y}{\lambda }$, or $y=u\lambda$. Hence

$$\frac{\mathrm{d}y}{\mathrm{d}\lambda }=\frac{\mathrm{d}u}{\mathrm{d}\lambda }\lambda +u$$

Applying the transformation $y=u\lambda$ to the above ODE in (1) gives

$$\begin{array}{rl}\frac{\mathrm{d}u}{\mathrm{d}\lambda }\lambda +u& =\frac{2u^2}{2u-1}\\ \frac{\mathrm{d}u}{\mathrm{d}\lambda }& =\frac{\frac{2u{\left(\lambda \right)}^2}{2u\left(\lambda \right)-1}-u\left(\lambda \right)}{\lambda }\end{array}$$

Or

$$u^{\prime }\left(\lambda \right)-\frac{\frac{2u{\left(\lambda \right)}^2}{2u\left(\lambda \right)-1}-u\left(\lambda \right)}{\lambda }=0$$

Or

$$2u^{\prime }\left(\lambda \right)\lambda u\left(\lambda \right)-u^{\prime }\left(\lambda \right)\lambda -u\left(\lambda \right)=0$$

Or

$$\lambda (2u\left(\lambda \right)-1)u^{\prime }\left(\lambda \right)-u\left(\lambda \right)=0$$

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

The ode

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

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }\left(\lambda \right)& =\frac{u\left(\lambda \right)}{\lambda (2u\left(\lambda \right)-1)}\\ & =f(\lambda )g(u)\end{array}$$

Where

$$\begin{array}{rl}f(\lambda )& =\frac{1}{\lambda }\\ g(u)& =\frac{u}{2u-1}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(\lambda )d\lambda \\ \int \frac{2u-1}{u}du& =\int \frac{1}{\lambda }d\lambda \end{array}$$

$$2u\left(\lambda \right)+\ln \left(\frac{1}{u\left(\lambda \right)}\right)=\ln \left(\lambda \right)+c_1$$

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

$$\frac{u}{2u-1}=0$$

for $u\left(\lambda \right)$ gives

$$\begin{array}{rl}u\left(\lambda \right)& =0\end{array}$$

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

$$\begin{array}{rl}2u\left(\lambda \right)+\ln \left(\frac{1}{u\left(\lambda \right)}\right)& =\ln \left(\lambda \right)+c_1\\ u\left(\lambda \right)& =0\end{array}$$

Converting $2u\left(\lambda \right)+\ln \left(\frac{1}{u\left(\lambda \right)}\right)=\ln \left(\lambda \right)+c_1$ back to $y$ gives

$$\begin{array}{r}\frac{\ln \left(\frac{\lambda }{y}\right)\lambda +2y}{\lambda }=\ln \left(\lambda \right)+c_1\end{array}$$

Converting $u\left(\lambda \right)=0$ back to $y$ gives

$$\begin{array}{r}y=0\end{array}$$

Solving for $y$ gives

$$\begin{array}{rl}y& =0\\ y& =\frac{c_1\lambda }{2}+\frac{\ln \left(\lambda \right)\lambda }{2}-\frac{\ln (-\frac{2}{\mathrm{LambertW}(-\frac{2{\mathrm{e}}^{-c_1}}{\lambda })})\lambda }{2}\end{array}$$

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

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

Eliminating $\lambda$ gives the solution for $y$.

$$\mathrm{RootOf}(x{\mathit{\text{\_Z}}}^2-1)\ln \left(\mathrm{RootOf}(x{\mathit{\text{\_Z}}}^2-1)y\left(x\right)\right)-\mathrm{RootOf}(x{\mathit{\text{\_Z}}}^2-1)\ln (-\frac{2}{\mathrm{LambertW}(-\frac{2{\mathrm{e}}^{-c_1}}{\mathrm{RootOf}(x{\mathit{\text{\_Z}}}^2-1)y\left(x\right)})})+\mathrm{RootOf}(x{\mathit{\text{\_Z}}}^2-1)c_1-2$$

Which can be written as

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

Solving for $y\left(x\right)$ gives

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

Summary of solutions found

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

✓ Maple. Time used: 0.040 (sec). Leaf size: 27

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

$$\begin{array}{rl}y& =0\\ y& =c_1{\mathrm{e}}^{-2\sqrt{x}}\\ y& =c_1{\mathrm{e}}^{2\sqrt{x}}\end{array}$$

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}& & \text{Let’s solve}\\ & & {\left(\frac{d}{dx}y\left(x\right)\right)}^2=\frac{y{\left(x\right)}^2}{x}\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & [\frac{d}{dx}y\left(x\right)=\frac{y\left(x\right)}{\sqrt{x}},\frac{d}{dx}y\left(x\right)=-\frac{y\left(x\right)}{\sqrt{x}}]\\ ◻& & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=\frac{y\left(x\right)}{\sqrt{x}}\\ & \circ & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{y\left(x\right)}=\frac{1}{\sqrt{x}}\\ & \circ & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{y\left(x\right)}dx=\int \frac{1}{\sqrt{x}}dx+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & \ln \left(y\left(x\right)\right)=2\sqrt{x}+\mathit{\text{\_C1}}\\ & \circ & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)={\mathrm{e}}^{2\sqrt{x}+\mathit{\text{\_C1}}}\\ & \circ & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\mathit{\text{\_C1}}{\mathrm{e}}^{2\sqrt{x}}\\ ◻& & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=-\frac{y\left(x\right)}{\sqrt{x}}\\ & \circ & \text{Separate variables}\\ & & \frac{\frac{d}{dx}y\left(x\right)}{y\left(x\right)}=-\frac{1}{\sqrt{x}}\\ & \circ & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}y\left(x\right)}{y\left(x\right)}dx=\int -\frac{1}{\sqrt{x}}dx+\mathit{\text{\_C1}}\\ & \circ & \text{Evaluate integral}\\ & & \ln \left(y\left(x\right)\right)=-2\sqrt{x}+\mathit{\text{\_C1}}\\ & \circ & \text{Solve for}y\left(x\right)\\ & & y\left(x\right)={\mathrm{e}}^{-2\sqrt{x}+\mathit{\text{\_C1}}}\\ & \circ & \text{Redefine the integration constant(s)}\\ & & y\left(x\right)=\mathit{\text{\_C1}}{\mathrm{e}}^{-2\sqrt{x}}\\ \bullet & & \text{Set of solutions}\\ & & \{y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-2\sqrt{x}},y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{2\sqrt{x}}\}\end{array}$$

✓ Mathematica. Time used: 0.069 (sec). Leaf size: 38

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

$$\begin{array}{r}y(x)\to c_1{e}^{-2\sqrt{x}}\\ y(x)\to c_1{e}^{2\sqrt{x}}\\ y(x)\to 0\end{array}$$

✓ Sympy. Time used: 0.427 (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)
 

$$[y\left(x\right)=C_1{e}^{-2x\sqrt{\frac{1}{x}}},y\left(x\right)=C_1{e}^{2x\sqrt{\frac{1}{x}}}]$$