2.1.45 Problem 45

Internal problem ID [8757]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 45
Date solved : Sunday, March 30, 2025 at 01:30:31 PM
CAS classification : [_dAlembert]

Solved using first_order_ode_dAlembert

Time used: 0.141 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }y& =1-x{y^{\prime }}^3\end{array}$$

Let $p=y^{\prime }$ the ode becomes

$$\begin{array}{r}py=-x{p}^3+1\end{array}$$

Solving for $y$ from the above results in

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

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. The above ode is dAlembert ode which is now solved.

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

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =-p^2\\ g& =\frac{1}{p}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p^2+p=(-2xp-\frac{1}{p^2})p^{\prime }\left(x\right)\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}{p}^2+p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =-1\\ p_2& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=-x-1\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

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

Inverting the above ode gives

$$\begin{array}{}\text{(4)}& \frac{d}{dp}x\left(p\right)=\frac{-2x\left(p\right)p-\frac{1}{p^2}}{p^2+p}\end{array}$$

This ODE is now solved for $x\left(p\right)$. The integrating factor is

$$\begin{array}{rl}\mu & ={\mathrm{e}}^{\int \frac{2}{p+1}dp}\\ \mu & ={(p+1)}^2\\ \text{(5)}& \mu & ={(p+1)}^2\end{array}$$

Integrating gives

$$\begin{array}{rl}x\left(p\right)& =\frac{1}{\mu }(\int \mu (-\frac{1}{p^3(p+1)})dp+c_1)\\ & =\frac{1}{\mu }(\frac{\frac{1}{2p^2}+\frac{1}{p}+c_1}{{(p+1)}^2}+c_1)\\ \text{(5)}& & =\frac{\frac{1}{2p^2}+\frac{1}{p}+c_1}{{(p+1)}^2}\end{array}$$

Now we need to eliminate $p$ between the above solution and (1A). The first method is to solve for $p$ from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for $p$ from Eq. (5) and substitute the result into (1A).

Eliminating $p$ from the following two equations

$$\begin{array}{rl}x& =\frac{\frac{1}{2p^2}+\frac{1}{p}+c_1}{{(p+1)}^2}\\ y& =-p^{2}x+\frac{1}{p}\end{array}$$

results in

$$\begin{array}{rl}p& =\mathrm{RootOf}(-1+2x{\mathit{\text{\_Z}}}^4+4x{\mathit{\text{\_Z}}}^3+(-2c_1+2x){\mathit{\text{\_Z}}}^2-2\mathit{\text{\_Z}})\end{array}$$

Substituting the above into Eq (1A) and simplifying gives

$$\begin{array}{rl}y& =-\frac{x\mathrm{RootOf}{(-1+2x{\mathit{\text{\_Z}}}^4+4x{\mathit{\text{\_Z}}}^3+(-2c_1+2x){\mathit{\text{\_Z}}}^2-2\mathit{\text{\_Z}})}^3-1}{\mathrm{RootOf}(-1+2x{\mathit{\text{\_Z}}}^4+4x{\mathit{\text{\_Z}}}^3+(-2c_1+2x){\mathit{\text{\_Z}}}^2-2\mathit{\text{\_Z}})}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =-\frac{x\mathrm{RootOf}{(-1+2x{\mathit{\text{\_Z}}}^4+4x{\mathit{\text{\_Z}}}^3+(-2c_1+2x){\mathit{\text{\_Z}}}^2-2\mathit{\text{\_Z}})}^3-1}{\mathrm{RootOf}(-1+2x{\mathit{\text{\_Z}}}^4+4x{\mathit{\text{\_Z}}}^3+(-2c_1+2x){\mathit{\text{\_Z}}}^2-2\mathit{\text{\_Z}})}\\ y& =-x-1\end{array}$$

✓ Maple. Time used: 0.072 (sec). Leaf size: 1791

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

$$\begin{array}{r}\text{Solution too large to show}\end{array}$$

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 
trying dAlembert 
<- dAlembert successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \left(\frac{d}{dx}y\left(x\right)\right)y\left(x\right)=1-x{\left(\frac{d}{dx}y\left(x\right)\right)}^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{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{6x}-\frac{2y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}},\frac{d}{dx}y\left(x\right)=-\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{12x}+\frac{y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}-\frac{\mathrm{I}\sqrt{3}(\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{6x}+\frac{2y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}})}{2},\frac{d}{dx}y\left(x\right)=-\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{12x}+\frac{y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}+\frac{\mathrm{I}\sqrt{3}(\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{6x}+\frac{2y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}})}{2}]\\ \bullet & & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{6x}-\frac{2y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}\\ \bullet & & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=-\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{12x}+\frac{y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}-\frac{\mathrm{I}\sqrt{3}(\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{6x}+\frac{2y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}})}{2}\\ \bullet & & \text{Solve the equation}\frac{d}{dx}y\left(x\right)=-\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{12x}+\frac{y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}+\frac{\mathrm{I}\sqrt{3}(\frac{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}}{6x}+\frac{2y\left(x\right)}{{\left((12\sqrt{3}\sqrt{\frac{4y{\left(x\right)}^3+27x}{x}}+108)x^2\right)}^{1/3}})}{2}\\ \bullet & & \text{Set of solutions}\\ & & \{\mathit{workingODE},\mathit{workingODE},\mathit{workingODE}\}\end{array}$$

✓ Mathematica. Time used: 86.107 (sec). Leaf size: 20717

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

Too large to display

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), x)**3 + y(x)*Derivative(y(x), x) - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out