8.27 problem 40

Internal problem ID [13054]
Internal file name [OUTPUT/11707_Wednesday_November_08_2023_03_29_03_AM_54714494/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. Review Exercises for chapter 1. page 136
Problem number: 40.
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]

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

8.27.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= y^{2}-2 y +1 \end {align*}

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

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

8.27.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}-2 y +1}d y &= \int {dt}\\ -\frac {1}{y -1}&= t +c_{1} \end {align*}

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

The solutions are \begin {align*} c_{1} = -1 \end {align*}

Trying the constant \begin {align*} c_{1} = -1 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {1}{y -1} = t -1 \end {align*}

The constant \(c_{1} = -1\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {1}{y-1} &= t -1 \\ \end{align*}

Verification of solutions

\[ -\frac {1}{y-1} = t -1 \] Verified OK.

8.27.3 Maple step by step solution

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

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: 0.016 (sec). Leaf size: 13

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

\[ y \left (t \right ) = \frac {t -2}{t -1} \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 14

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

\[ y(t)\to \frac {t-2}{t-1} \]