2.1.86 Internal
problem
ID
[10072]Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
85Date
solved
:
Saturday, November 29, 2025 at 03:28:43 PMCAS
classification
:
[_Bernoulli]
\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.
Time used: 0.155 (sec)
Solve
\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
\[ v \left (x \right ) = x +1+{\mathrm e}^{x} c_1 \]
The substitution
\(v = y^{1-n}\) is now
used to convert the above solution back to
\(y\) which results in
\[
\frac {1}{\sqrt {y}} = x +1+{\mathrm e}^{x} c_1
\]
Solving for initial conditions the
solution is
\begin{align*}
\frac {1}{\sqrt {y}} &= x +1 \\
\end{align*}
Figure 2.176: Slope field \(y^{\prime } = 2 y \left (x \sqrt {y}-1\right )\) Summary of solutions found
\begin{align*}
\frac {1}{\sqrt {y}} &= x +1 \\
\end{align*}
✓ Time used: 0.039 (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
✓ 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*}
✗ 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