Internal problem ID [14192]
Internal file name [OUTPUT/13873_Saturday_March_09_2024_03_56_34_PM_782895/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.2, page 39
Problem number: 35.
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]
\[ \boxed {y^{\prime }-y^{3}=1} \]
Integrating both sides gives \begin {align*} \int \frac {1}{y^{3}+1}d y &= \int {dt}\\ \int _{}^{y}\frac {1}{\textit {\_a}^{3}+1}d \textit {\_a}&= t +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\textit {\_a}^{3}+1}d \textit {\_a} &= t +c_{1} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\textit {\_a}^{3}+1}d \textit {\_a} = t +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{3}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{3}+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{3}+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{3}+1}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y+1\right )}{3}-\frac {\ln \left (y^{2}-y+1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 y-1\right ) \sqrt {3}}{3}\right )}{3}=t +c_{1} \\ \bullet & {} & \textrm {Convert}\hspace {3pt} \arctan \mathrm {to \esapos ln\esapos } \\ {} & {} & \frac {\ln \left (y+1\right )}{3}-\frac {\ln \left (y^{2}-y+1\right )}{6}+\frac {\mathrm {I} \sqrt {3}\, \left (\ln \left (1+\frac {\mathrm {I} \left (1-2 y\right ) \sqrt {3}}{3}\right )-\ln \left (1+\frac {\mathrm {I} \left (2 y-1\right ) \sqrt {3}}{3}\right )\right )}{6}=t +c_{1} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable <- separable successful`
✓ Solution by Maple
Time used: 0.563 (sec). Leaf size: 51
dsolve(diff(y(t),t)=y(t)^3+1,y(t), singsol=all)
\[ y \left (t \right ) = \frac {1}{2}+\frac {\sqrt {3}\, \tan \left (\operatorname {RootOf}\left (-\sqrt {3}\, \ln \left (\cos \left (\textit {\_Z} \right )^{2}\right )-2 \sqrt {3}\, \ln \left (\sqrt {3}+\tan \left (\textit {\_Z} \right )\right )+6 \sqrt {3}\, c_{1} +6 \sqrt {3}\, t -6 \textit {\_Z} \right )\right )}{2} \]
✓ Solution by Mathematica
Time used: 0.196 (sec). Leaf size: 80
DSolve[y'[t]==y[t]^3+1,y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to \text {InverseFunction}\left [-\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][t+c_1] \\ y(t)\to -1 \\ y(t)\to \sqrt [3]{-1} \\ y(t)\to -(-1)^{2/3} \\ \end{align*}