Internal problem ID [12988]
Internal file name [OUTPUT/11641_Tuesday_November_07_2023_11_53_56_PM_58776703/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 1.6 page 89
Problem number: 37 (viii).
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]
\[ \boxed {y^{\prime }-y^{2}+y^{3}=0} \]
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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}+y^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-y^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}-y^{3}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{2}-y^{3}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y}+\ln \left (y\right )-\ln \left (y-1\right )=t +c_{1} \end {array} \]
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: 0.109 (sec). Leaf size: 20
dsolve(diff(y(t),t)=y(t)^2-y(t)^3,y(t), singsol=all)
\[ y \left (t \right ) = \frac {1}{\operatorname {LambertW}\left (-\frac {{\mathrm e}^{-t -1}}{c_{1}}\right )+1} \]
✓ Solution by Mathematica
Time used: 0.408 (sec). Leaf size: 40
DSolve[y'[t]==y[t]^2-y[t]^3,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*}