2.1.49 Problem 49

Internal problem ID [10307]
Book : First order enumerated odes
Section : section 1
Problem number : 49
Date solved : Thursday, November 27, 2025 at 10:32:48 AM
CAS classification : [_quadrature]

Solved using first_order_ode_homog_type_G

Time used: 0.135 (sec)

Solve

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

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

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \sqrt {x}\\ &= \frac {x^{{3}/{2}}}{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 {3 x^{{3}/{2}}}{2 y}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {x^{{3}/{2}}}{y}\\ \alpha &= \frac {f_x}{f_y} \\ &=-{\frac {3}{2}} \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^{{3}/{2}}}} \end{align*}

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

\begin{align*} F \left (z \right ) &=\frac {1}{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

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {1}{z \left (\frac {3}{2}-\frac {1}{z}\right )}d z = 0 \]

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

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) -\sqrt {x}\\ &= -\frac {x^{{3}/{2}}}{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 {3 x^{{3}/{2}}}{2 y}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {x^{{3}/{2}}}{y}\\ \alpha &= \frac {f_x}{f_y} \\ &=-{\frac {3}{2}} \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^{{3}/{2}}}} \end{align*}

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

\begin{align*} F \left (z \right ) &=-\frac {1}{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

\[ \ln \left (x \right )-c_2 +\int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {1}{z \left (\frac {3}{2}+\frac {1}{z}\right )}d z = 0 \]

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {1}{z \left (\frac {3}{2}-\frac {1}{z}\right )}d z &= 0 \\ \ln \left (x \right )-c_2 +\int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {1}{z \left (\frac {3}{2}+\frac {1}{z}\right )}d z &= 0 \\ \end{align*}

Solved using first_order_ode_parametric method

Time used: 0.101 (sec)

Solve

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

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

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

Isolating \(x\) gives

\begin{align*} x = \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} } \\ &= 2 \lambda ^{2}\\ &= 2 \lambda ^{2} \end{align*}

Which is now solved for \(y\).

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

\begin{align*} \int {dy} &= \int {2 \lambda ^{2}\, d\lambda }\\ y \left (\lambda \right ) &= \frac {2 \lambda ^{3}}{3} + c_1 \end{align*}

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

\begin{align*} y &= \frac {2 \lambda ^{3}}{3}+c_1 \\ x &= \lambda ^{2} \\ \end{align*}

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

\begin{align*} y &= c_1 -\frac {2 x^{{3}/{2}}}{3} \\ y &= c_1 +\frac {2 x^{{3}/{2}}}{3} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_1 -\frac {2 x^{{3}/{2}}}{3} \\ y &= c_1 +\frac {2 x^{{3}/{2}}}{3} \\ \end{align*}

✓ Maple. Time used: 0.016 (sec). Leaf size: 21

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

\begin{align*} y &= \frac {2 x^{{3}/{2}}}{3}+c_1 \\ y &= -\frac {2 x^{{3}/{2}}}{3}+c_1 \\ \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 
<- differential order: 1; missing  y(x)  successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\sqrt {x}, \frac {d}{d x}y \left (x \right )=-\sqrt {x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\sqrt {x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int \sqrt {x}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {2 x^{{3}/{2}}}{3}+\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\sqrt {x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int -\sqrt {x}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y \left (x \right )=-\frac {2 x^{{3}/{2}}}{3}+\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=-\frac {2 x^{{3}/{2}}}{3}+\mathit {C1} , y \left (x \right )=\frac {2 x^{{3}/{2}}}{3}+\mathit {C1} \right \} \end {array} \]

✓ Mathematica. Time used: 0.003 (sec). Leaf size: 33

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

\begin{align*} y(x)&\to -\frac {2 x^{3/2}}{3}+c_1\\ y(x)&\to \frac {2 x^{3/2}}{3}+c_1 \end{align*}

✓ Sympy. Time used: 0.221 (sec). Leaf size: 24

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

\[ \left [ y{\left (x \right )} = C_{1} - \frac {2 x^{\frac {3}{2}}}{3}, \ y{\left (x \right )} = C_{1} + \frac {2 x^{\frac {3}{2}}}{3}\right ] \]