2.1.86 Problem 85

Internal problem ID [10072]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 85
Date solved : Monday, March 09, 2026 at 03:11:55 AM
CAS classification : [_Bernoulli]

2.1.86.1 Existence and uniqueness analysis

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

This is non linear first order ODE. In canonical form it is written as

\begin{align*} y^{\prime } &= f(x,y)\\ &= 2 y \left (x \sqrt {y}-1\right ) \end{align*}

The \(x\) domain of \(f(x,y)\) when \(y=1\) is

\[ \{-\infty <x <\infty \} \]

And the point \(x_0 = 0\) is inside this domain. The \(y\) domain of \(f(x,y)\) when \(x=0\) is

\[ \{-\infty <y <\infty \} \]

And the point \(y_0 = 1\) is inside this domain. Now we will look at the continuity of

\begin{align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (2 y \left (x \sqrt {y}-1\right )\right ) \\ &= -2+3 x \sqrt {y} \end{align*}

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=1\) is

\[ \{-\infty <x <\infty \} \]

And the point \(x_0 = 0\) is inside this domain. The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is

\[ \{-\infty <y <\infty \} \]

And the point \(y_0 = 1\) is inside this domain. Therefore solution exists and is unique.

2.1.86.2 Solved using first_order_ode_bernoulli

0.417 (sec)

Entering first order ode bernoulli solver

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

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= 2 y \left (x \sqrt {y}-1\right ) \end{align*}

This is a Bernoulli ODE.

\[ y' = \left (-2\right ) y + \left (2 x\right )y^{{3}/{2}} \tag {1} \]

The standard Bernoulli ODE has the form

\[ y' = f_0(x)y+f_1(x)y^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-2\\ f_1 &=2 x \end{align*}

The first step is to divide the above equation by \(y^n \) which gives

\[ \frac {y'}{y^n} = f_0(x) y^{1-n} +f_1(x) \tag {3} \]

The next step is use the substitution \(v = y^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(y(x)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(x)&=-2\\ f_1(x)&=2 x\\ n &={\frac {3}{2}} \end{align*}

Dividing both sides of ODE (1) by \(y^n=y^{{3}/{2}}\) gives

\begin{align*} y'\frac {1}{y^{{3}/{2}}} &= -\frac {2}{\sqrt {y}} +2 x \tag {4} \end{align*}

Let

\begin{align*} v &= y^{1-n} \\ &= \frac {1}{\sqrt {y}} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(x\) gives

\begin{align*} v' &= -\frac {1}{2 y^{{3}/{2}}}y' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -2 v^{\prime }\left (x \right )&= -2 v \left (x \right )+2 x\\ v' &= v -x \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (x \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} v^{\prime }\left (x \right ) + q(x)v \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-1\\ p(x) &=-x \end{align*}

The integrating factor \(\mu \) is

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

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \left (\mu \right ) \left (-x\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (v \,{\mathrm e}^{-x}\right ) &= \left ({\mathrm e}^{-x}\right ) \left (-x\right ) \\ \mathrm {d} \left (v \,{\mathrm e}^{-x}\right ) &= \left (-x \,{\mathrm e}^{-x}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{-x}&= \int {-x \,{\mathrm e}^{-x} \,dx} \\ &=\left (x +1\right ) {\mathrm e}^{-x} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-x}\) gives the final solution

\[ v \left (x \right ) = {\mathrm e}^{x} \left (\left (x +1\right ) {\mathrm e}^{-x}+c_1 \right ) \]

The substitution \(v = y^{1-n}\) is now used to convert the above solution back to \(y\) which results in

\[ \frac {1}{\sqrt {y}} = {\mathrm e}^{x} \left (\left (x +1\right ) {\mathrm e}^{-x}+c_1 \right ) \]

Simplifying the above gives

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

Solving for initial conditions the solution becomes

\begin{align*} \frac {1}{\sqrt {y}} &= x +1 \\ \end{align*}

Solving for \(y\) gives

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

Solution plot for \(y^{\prime } = 2 y \left (x \sqrt {y}-1\right )\)	Direction fields for \(y^{\prime } = 2 y \left (x \sqrt {y}-1\right )\)
Isoclines for \(y^{\prime } = 2 y \left (x \sqrt {y}-1\right )\)

Summary of solutions found

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

2.1.86.3 ✓ Maple. Time used: 0.035 (sec). Leaf size: 9

ode:=diff(y(x),x) = 2*y(x)*(x*y(x)^(1/2)-1); 
ic:=[y(0) = 1]; 
dsolve([ode,op(ic)],y(x), singsol=all);
 

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

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful
 

2.1.86.4 ✓ Mathematica. Time used: 0.425 (sec). Leaf size: 20

ode=D[y[x],x]==2*y[x]*(x*Sqrt[y[x]-1]); 
ic={y[0]==1}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to 1\\ y(x)&\to \sec ^2\left (\frac {x^2}{2}\right ) \end{align*}

2.1.86.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((-2*x*sqrt(y(x)) + 2)*y(x) + Derivative(y(x), x),0) 
ics = {y(0): 1} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : Initial conditions produced too many solutions for constants
 

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)) 
 
('factorable', 'Bernoulli', '1st_power_series', 'lie_group', 'Bernoulli_Integral')