2.1.20 Problem 20 Internal
problem
ID
[10006] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
20 Date
solved
:
Monday, December 08, 2025 at 07:02:57 PMCAS
classification
:
[_Bernoulli]
2.1.20.1 Solved using first_order_ode_bernoulli 0.167 (sec)
Entering first order ode bernoulli solver
\begin{align*}
y^{\prime }+\frac {y}{3}&=\frac {\left (1-2 x \right ) y^{4}}{3} \\
\end{align*}
In canonical form, the ODE is
\begin{align*} y' &= F(x,y)\\ &= -\frac {1}{3} y -\frac {2}{3} y^{4} x +\frac {1}{3} y^{4} \end{align*}
This is a Bernoulli ODE.
\[ y' = \left (-{\frac {1}{3}}\right ) y + \left (-\frac {2 x}{3}+\frac {1}{3}\right )y^{4} \tag {1} \]
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 &=-{\frac {1}{3}}\\ f_1 &=-\frac {2 x}{3}+\frac {1}{3} \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)&=-{\frac {1}{3}}\\ f_1(x)&=-\frac {2 x}{3}+\frac {1}{3}\\ n &=4 \end{align*}
Dividing both sides of ODE (1) by \(y^n=y^{4}\) gives
\begin{align*} y'\frac {1}{y^{4}} &= -\frac {1}{3 y^{3}} -\frac {2 x}{3}+\frac {1}{3} \tag {4} \end{align*}
Let
\begin{align*} v &= y^{1-n} \\ &= \frac {1}{y^{3}} \tag {5} \end{align*}
Taking derivative of equation (5) w.r.t \(x\) gives
\begin{align*} v' &= -\frac {3}{y^{4}}y' \tag {6} \end{align*}
Substituting equations (5) and (6) into equation (4) gives
\begin{align*} -\frac {v^{\prime }\left (x \right )}{3}&= -\frac {v \left (x \right )}{3}-\frac {2 x}{3}+\frac {1}{3}\\ v' &= v +2 x -1 \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) &=2 x -1 \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 (2 x -1\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (v \,{\mathrm e}^{-x}\right ) &= \left ({\mathrm e}^{-x}\right ) \left (2 x -1\right ) \\
\mathrm {d} \left (v \,{\mathrm e}^{-x}\right ) &= \left (\left (2 x -1\right ) {\mathrm e}^{-x}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives \begin{align*} v \,{\mathrm e}^{-x}&= \int {\left (2 x -1\right ) {\mathrm e}^{-x} \,dx} \\ &=\left (-2 x -1\right ) {\mathrm e}^{-x} + c_1 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{-x}\) gives the final solution
\[ v \left (x \right ) = c_1 \,{\mathrm e}^{x}-2 x -1 \]
The substitution \(v = y^{1-n}\) is now
used to convert the above solution back to \(y\) which results in \[
\frac {1}{y^{3}} = c_1 \,{\mathrm e}^{x}-2 x -1
\]
Figure 2.53: Slope field \(y^{\prime }+\frac {y}{3} = \frac {\left (1-2 x \right ) y^{4}}{3}\) Summary of solutions found
\begin{align*}
\frac {1}{y^{3}} &= c_1 \,{\mathrm e}^{x}-2 x -1 \\
\end{align*}
2.1.20.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 63 ode := diff ( y ( x ), x )+1/3* y ( x ) = 1/3*(1-2*x)*y(x)^4;
dsolve ( ode , y ( x ), singsol=all);
\begin{align*}
y &= \frac {1}{\left ({\mathrm e}^{x} c_1 -2 x -1\right )^{{1}/{3}}} \\
y &= -\frac {1+i \sqrt {3}}{2 \left ({\mathrm e}^{x} c_1 -2 x -1\right )^{{1}/{3}}} \\
y &= \frac {i \sqrt {3}-1}{2 \left ({\mathrm e}^{x} c_1 -2 x -1\right )^{{1}/{3}}} \\
\end{align*}
Maple trace
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+\frac {y \left (x \right )}{3}=\frac {\left (1-2 x \right ) y \left (x \right )^{4}}{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{3}+\frac {\left (1-2 x \right ) y \left (x \right )^{4}}{3} \end {array} \]
2.1.20.3 ✓ Mathematica. Time used: 8.39 (sec). Leaf size: 139 ode = D [ y [ x ], x ]+ y [ x ]/3== (1-2*x)/3*y[x]^4;
ic ={}; DSolve [{ ode , ic }, y [ x ], x , IncludeSingularSolutions -> True ] \begin{align*} y(x)&\to \frac {1}{\sqrt [3]{e^x \left (-3 \int _1^x\frac {1}{3} e^{-K[1]} (1-2 K[1])dK[1]+c_1\right )}}\\ y(x)&\to -\frac {\sqrt [3]{-1}}{\sqrt [3]{e^x \left (-3 \int _1^x\frac {1}{3} e^{-K[1]} (1-2 K[1])dK[1]+c_1\right )}}\\ y(x)&\to \frac {(-1)^{2/3}}{\sqrt [3]{e^x \left (-3 \int _1^x\frac {1}{3} e^{-K[1]} (1-2 K[1])dK[1]+c_1\right )}}\\ y(x)&\to 0 \end{align*}
2.1.20.4 ✓ Sympy. Time used: 1.584 (sec). Leaf size: 78 from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((2*x - 1)*y(x)**4/3 + y(x)/3 + Derivative(y(x), x),0)
ics = {}
dsolve ( ode , func = y ( x ), ics = ics ) \[
\left [ y{\left (x \right )} = \sqrt [3]{- \frac {1}{C_{1} e^{x} + 2 x + 1}}, \ y{\left (x \right )} = \frac {\sqrt [3]{- \frac {1}{C_{1} e^{x} + 2 x + 1}} \left (-1 - \sqrt {3} i\right )}{2}, \ y{\left (x \right )} = \frac {\sqrt [3]{- \frac {1}{C_{1} e^{x} + 2 x + 1}} \left (-1 + \sqrt {3} i\right )}{2}\right ]
\]