Internal problem ID [13289]
Internal file name [OUTPUT/12461_Wednesday_February_14_2024_02_06_24_AM_81205944/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 3. Some basics about First order equations. Additional exercises. page
63
Problem number: 3.4 c.
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]
\[ \boxed {y^{\prime }-y^{3}=8} \]
Integrating both sides gives \begin {align*} \int \frac {1}{y^{3}+8}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\textit {\_a}^{3}+8}d \textit {\_a}&= x +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\textit {\_a}^{3}+8}d \textit {\_a} &= x +c_{1} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\textit {\_a}^{3}+8}d \textit {\_a} = x +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{3}=8 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{3}+8 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{3}+8}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{3}+8}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y+2\right )}{12}-\frac {\ln \left (y^{2}-2 y+4\right )}{24}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 y-2\right ) \sqrt {3}}{6}\right )}{12}=x +c_{1} \\ \bullet & {} & \textrm {Convert}\hspace {3pt} \arctan \mathrm {to \esapos ln\esapos } \\ {} & {} & \frac {\ln \left (y+2\right )}{12}-\frac {\ln \left (y^{2}-2 y+4\right )}{24}+\frac {\mathrm {I} \sqrt {3}\, \left (\ln \left (1+\frac {\mathrm {I} \left (-y+1\right ) \sqrt {3}}{3}\right )-\ln \left (1+\frac {\mathrm {I} \left (y-1\right ) \sqrt {3}}{3}\right )\right )}{24}=x +c_{1} \end {array} \]
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: 0.516 (sec). Leaf size: 50
dsolve(diff(y(x),x)-y(x)^3=8,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {3}\, \tan \left (\operatorname {RootOf}\left (-\sqrt {3}\, \ln \left (\cos \left (\textit {\_Z} \right )^{2}\right )-2 \sqrt {3}\, \ln \left (\sqrt {3}+\tan \left (\textit {\_Z} \right )\right )+24 \sqrt {3}\, c_{1} +24 \sqrt {3}\, x -6 \textit {\_Z} \right )\right )+1 \]
✓ Solution by Mathematica
Time used: 0.206 (sec). Leaf size: 83
DSolve[y'[x]-y[x]^3==8,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {1}{24} \log \left (\text {$\#$1}^2-2 \text {$\#$1}+4\right )+\frac {\arctan \left (\frac {\text {$\#$1}-1}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{12} \log (\text {$\#$1}+2)\&\right ][x+c_1] \\ y(x)\to -2 \\ y(x)\to 2 \sqrt [3]{-1} \\ y(x)\to -2 (-1)^{2/3} \\ \end{align*}