Internal problem ID [12916]
Internal file name [OUTPUT/11569_Tuesday_November_07_2023_11_27_12_PM_89430785/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.3 page 47
Problem number: 16 (iii).
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]
\[ \boxed {y^{\prime }-y^{3}-y^{2}=0} \]
Integrating both sides gives \begin {align*} \int \frac {1}{y^{3}+y^{2}}d y &= \int {dt}\\ \int _{}^{y}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}}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}+\textit {\_a}^{2}}d \textit {\_a} &= t +c_{1} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}}d \textit {\_a} = t +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{3}-y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{3}+y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{3}+y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{3}+y^{2}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y}-\ln \left (y\right )+\ln \left (y+1\right )=t +c_{1} \end {array} \]
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: 0.172 (sec). Leaf size: 18
dsolve(diff(y(t),t)=y(t)^3+y(t)^2,y(t), singsol=all)
\[ y \left (t \right ) = -\frac {1}{\operatorname {LambertW}\left (-c_{1} {\mathrm e}^{t -1}\right )+1} \]
✓ Solution by Mathematica
Time used: 0.318 (sec). Leaf size: 38
DSolve[y'[t]==y[t]^3+y[t]^2,y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to \text {InverseFunction}\left [-\frac {1}{\text {$\#$1}}-\log (\text {$\#$1})+\log (\text {$\#$1}+1)\&\right ][t+c_1] \\ y(t)\to -1 \\ y(t)\to 0 \\ \end{align*}