Internal problem ID [14195]
Internal file name [OUTPUT/13876_Saturday_March_09_2024_03_56_34_PM_13684374/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: 38.
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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{3}+y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{3}-y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{3}-y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{3}-y^{2}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y-1\right )+\frac {1}{y}-\ln \left (y\right )=t +c_{1} \end {array} \]
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: 16
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: 0.225 (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 {$\#$1}}+\log (1-\text {$\#$1})-\log (\text {$\#$1})\&\right ][t+c_1] \\ y(t)\to 0 \\ y(t)\to 1 \\ \end{align*}