2.14 problem 15 b(4)

Internal problem ID [12912]
Internal file name [OUTPUT/11565_Tuesday_November_07_2023_11_27_09_PM_69487583/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(4).
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 {3}{2}}\right ] \end {align*}

2.14.1 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

2.14.2 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 \(S={\frac {3}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -2+\ln \left (3\right ) = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -2+\ln \left (3\right ) \end {align*}

Trying the constant \begin {align*} c_{1} = -2+\ln \left (3\right ) \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 ) = t -2+\ln \left (3\right ) \end {align*}

The constant \(c_{1} = -2+\ln \left (3\right )\) 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 ) &= t -2+\ln \left (3\right ) \\ \end{align*}

Verification of solutions

\[ -\frac {1}{S-1}-\ln \left (S-1\right )+\ln \left (S\right ) = t -2+\ln \left (3\right ) \] Verified OK.

2.14.3 Maple step by step solution

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

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: 11.64 (sec). Leaf size: 41

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

\[ S \left (t \right ) = {\mathrm e}^{\operatorname {RootOf}\left (-\ln \left ({\mathrm e}^{\textit {\_Z}}+1\right ) {\mathrm e}^{\textit {\_Z}}+{\mathrm e}^{\textit {\_Z}} \ln \left (3\right )+\textit {\_Z} \,{\mathrm e}^{\textit {\_Z}}+t \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+1\right )}+1 \]

✓ Solution by Mathematica

Time used: 0.885 (sec). Leaf size: 31

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

\[ S(t)\to \text {InverseFunction}\left [-\frac {1}{\text {$\#$1}-1}-\log (\text {$\#$1}-1)+\log (\text {$\#$1})\&\right ][t-2+\log (3)] \]