2.1.20 Problem 20

Internal problem ID [8732]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 20
Date solved : Sunday, March 30, 2025 at 01:28:24 PM
CAS classification : [_Bernoulli]

Solved using first_order_ode_bernoulli

Time used: 0.187 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }+\frac{y}{3}& =\frac{(1-2x)y^4}{3}\end{array}$$

In canonical form, the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(x,y)\\ & =-\frac{1}{3}y-\frac{2}{3}{y}^{4}x+\frac{1}{3}{y}^4\end{array}$$

This is a Bernoulli ODE.

$$\begin{array}{}\text{(1)}& y^{\prime }=(-\frac{1}{3})y+(-\frac{2x}{3}+\frac{1}{3})y^4\end{array}$$

The standard Bernoulli ODE has the form

$$\begin{array}{}\text{(2)}& y^{\prime }=f_0(x)y+f_1(x)y^n\end{array}$$

Comparing this to (1) shows that

$$\begin{array}{rl}{f}_0& =-\frac{1}{3}\\ f_1& =-\frac{2x}{3}+\frac{1}{3}\end{array}$$

The first step is to divide the above equation by $y^n$ which gives

$$\begin{array}{}\text{(3)}& \frac{y^{\prime }}{y^n}=f_0(x)y^{1-n}+f_1(x)\end{array}$$

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

$$\begin{array}{rl}{f}_0(x)& =-\frac{1}{3}\\ f_1(x)& =-\frac{2x}{3}+\frac{1}{3}\\ n& =4\end{array}$$

Dividing both sides of ODE (1) by $y^n=y^4$ gives

$$\begin{array}{}\text{(4)}& y^{\prime }\frac{1}{y^4}& =-\frac{1}{3y^3}-\frac{2x}{3}+\frac{1}{3}\end{array}$$

Let

$$\begin{array}{rl}v& =y^{1-n}\\ \text{(5)}& & =\frac{1}{y^3}\end{array}$$

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

$$\begin{array}{}\text{(6)}& v^{\prime }& =-\frac{3}{y^4}{y}^{\prime }\end{array}$$

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

$$\begin{array}{rl}-\frac{v^{\prime }\left(x\right)}{3}& =-\frac{v\left(x\right)}{3}-\frac{2x}{3}+\frac{1}{3}\\ \text{(7)}& v^{\prime }& =v+2x-1\end{array}$$

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

In canonical form a linear first order is

$$\begin{array}{rl}{v}^{\prime }\left(x\right)+q(x)v\left(x\right)& =p(x)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(x)& =-1\\ p(x)& =2x-1\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int (-1)dx}\\ & ={\mathrm{e}}^{-x}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu v\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu v\right)& =\left(\mu \right)(2x-1)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(v{\mathrm{e}}^{-x}\right)& =\left({\mathrm{e}}^{-x}\right)(2x-1)\\ \mathrm{d}\left(v{\mathrm{e}}^{-x}\right)& =\left((2x-1){\mathrm{e}}^{-x}\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}v{\mathrm{e}}^{-x}& =\int (2x-1){\mathrm{e}}^{-x}dx\\ & =(-2x-1){\mathrm{e}}^{-x}+c_1\end{array}$$

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

$$v\left(x\right)={\mathrm{e}}^x{c}_1-2x-1$$

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

$$\frac{1}{y^3}={\mathrm{e}}^x{c}_1-2x-1$$

Solving for $y$ gives

$$\begin{array}{rl}y& =\frac{1}{{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =-\frac{1}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}-\frac{i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =-\frac{1}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}+\frac{i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\end{array}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{1}{{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =-\frac{1+i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =\frac{-1+i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\end{array}$$

Figure 2.44: Slope field $y^{\prime }+\frac{y}{3}=\frac{(1-2x)y^4}{3}$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{1}{{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =-\frac{1+i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =\frac{-1+i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 61

ode:=diff(y(x),x)+1/3*y(x) = 1/3*(-2*x+1)*y(x)^4; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}y& =\frac{1}{{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =-\frac{1+i\sqrt{3}}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\\ y& =\frac{i\sqrt{3}-1}{2{({\mathrm{e}}^x{c}_1-2x-1)}^{1/3}}\end{array}$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}y\left(x\right)+\frac{y\left(x\right)}{3}=\frac{(1-2x)y{\left(x\right)}^4}{3}\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=-\frac{y\left(x\right)}{3}+\frac{(1-2x)y{\left(x\right)}^4}{3}\end{array}$$

✓ Mathematica. Time used: 5.413 (sec). Leaf size: 76

ode=D[y[x],x]+y[x]/3== (1-2*x)/3*y[x]^4; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{1}{\sqrt[3]{-2x+c_1{e}^x-1}}\\ y(x)\to -\frac{\sqrt[3]{-1}}{\sqrt[3]{-2x+c_1{e}^x-1}}\\ y(x)\to \frac{(-1{)}^{2/3}}{\sqrt[3]{-2x+c_1{e}^x-1}}\\ y(x)\to 0\end{array}$$

✓ Sympy. Time used: 2.458 (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)
 

$$[y\left(x\right)=\sqrt[3]{-\frac{1}{C_1{e}^x+2x+1}},y\left(x\right)=\frac{\sqrt[3]{-\frac{1}{C_1{e}^x+2x+1}}(-1-\sqrt{3}i)}{2},y\left(x\right)=\frac{\sqrt[3]{-\frac{1}{C_1{e}^x+2x+1}}(-1+\sqrt{3}i)}{2}]$$