2.11 problem 15 b(1)

Internal problem ID [12909]
Internal file name [OUTPUT/11562_Tuesday_November_07_2023_11_27_05_PM_28153444/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: 15 b(1).
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]

\[ \boxed {S^{\prime }-S^{3}+2 S^{2}-S=0} \] With initial conditions \begin {align*} \left [S \left (0\right ) = {\frac {1}{2}}\right ] \end {align*}

2.11.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} S^{\prime } &= f(t,S)\\ &= S^{3}-2 S^{2}+S \end {align*}

The \(S\) domain of \(f(t,S)\) when \(t=0\) is \[ \{-\infty

The \(S\) domain of \(\frac {\partial f}{\partial S}\) when \(t=0\) is \[ \{-\infty

2.11.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{S^{3}-2 S^{2}+S}d S &= \int {dt}\\ -\frac {1}{S -1}-\ln \left (S -1\right )+\ln \left (S \right )&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(S={\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -i \pi +2 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -i \pi +2 \end {align*}

Trying the constant \begin {align*} c_{1} = -i \pi +2 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {1}{S -1}-\ln \left (S -1\right )+\ln \left (S \right ) = -i \pi +t +2 \end {align*}

The constant \(c_{1} = -i \pi +2\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {1}{S-1}-\ln \left (S-1\right )+\ln \left (S\right ) &= -i \pi +t +2 \\ \end{align*}

Verification of solutions

\[ -\frac {1}{S-1}-\ln \left (S-1\right )+\ln \left (S\right ) = -i \pi +t +2 \] Verified OK.

2.11.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [S^{\prime }-S^{3}+2 S^{2}-S=0, S \left (0\right )=\frac {1}{2}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & S^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & S^{\prime }=S^{3}-2 S^{2}+S \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {S^{\prime }}{S^{3}-2 S^{2}+S}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {S^{\prime }}{S^{3}-2 S^{2}+S}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{S-1}-\ln \left (S-1\right )+\ln \left (S\right )=t +c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} S \left (0\right )=\frac {1}{2} \\ {} & {} & -\mathrm {I} \pi +2=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\mathrm {I} \pi +2 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\mathrm {I} \pi +2\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & -\frac {1}{S-1}-\ln \left (S-1\right )+\ln \left (S\right )=t -\mathrm {I} \pi +2 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & -\frac {1}{S-1}-\ln \left (S-1\right )+\ln \left (S\right )=t -\mathrm {I} \pi +2 \end {array} \]

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: 1.39 (sec). Leaf size: 37

dsolve([diff(S(t),t)=S(t)^3-2*S(t)^2+S(t),S(0) = 1/2],S(t), singsol=all)
 

\[ S \left (t \right ) = {\mathrm e}^{\operatorname {RootOf}\left (-i \pi \,{\mathrm e}^{\textit {\_Z}}-\ln \left ({\mathrm e}^{\textit {\_Z}}+1\right ) {\mathrm e}^{\textit {\_Z}}+\textit {\_Z} \,{\mathrm e}^{\textit {\_Z}}+t \,{\mathrm e}^{\textit {\_Z}}+2 \,{\mathrm e}^{\textit {\_Z}}+1\right )}+1 \]

✗ Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[{S'[t]==S[t]^3-2*S[t]^2+S[t],{S[0]==1/2}},S[t],t,IncludeSingularSolutions -> True]
 

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