1.23 problem 2(i)

1.23.1 Solving as quadrature ode
1.23.2 Maple step by step solution

Internal problem ID [6127]
Internal file name [OUTPUT/5375_Sunday_June_05_2022_03_35_41_PM_50895992/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.2 THE NATURE OF SOLUTIONS. Page 9
Problem number: 2(i).
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "quadrature"

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {\left (x^{3}+1\right ) y^{\prime }=x} \]

1.23.1 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { \frac {x}{x^{3}+1}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}+c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}+c_{1} \\ \end{align*}

Figure 40: Slope field plot

Verification of solutions

\[ y = -\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}+c_{1} \] Verified OK.

1.23.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{3}+1\right ) y^{\prime }=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x}{x^{3}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {x}{x^{3}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (1+x \right )}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (1+x \right )}{3}+c_{1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 39

dsolve((1+x^3)*diff(y(x),x)=x,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (x +1\right )}{3}+c_{1} \]

Solution by Mathematica

Time used: 0.01 (sec). Leaf size: 48

DSolve[(1+x^3)*y'[x]==x,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{6} \left (2 \sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )+\log \left (x^2-x+1\right )-2 \log (x+1)+6 c_1\right ) \]