6.7 problem 7

Internal problem ID [12676]
Internal file name [OUTPUT/11329_Friday_November_03_2023_06_30_29_AM_99627193/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.3.2, page 63
Problem number: 7.
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]

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

6.7.1 Existence and uniqueness analysis

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

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

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

6.7.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}-2 y}d y &= \int {dx}\\ \frac {\ln \left (y -2\right )}{2}-\frac {\ln \left (y \right )}{2}&= x +c_{1} \end {align*}

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

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

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

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

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

Summary

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

Verification of solutions

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

6.7.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+2 y-y^{2}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 y+y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-2 y+y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-2 y+y^{2}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y-2\right )}{2}-\frac {\ln \left (y\right )}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {2}{{\mathrm e}^{2 x +2 c_{1}}-1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {2}{{\mathrm e}^{2 c_{1}}-1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\mathrm {I}}{2} \pi \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\mathrm {I}}{2} \pi \hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {2}{{\mathrm e}^{2 x}+1} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2}{{\mathrm e}^{2 x}+1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 14

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

\[ y \left (x \right ) = \frac {2}{{\mathrm e}^{2 x}+1} \]

✓ Solution by Mathematica

Time used: 0.013 (sec). Leaf size: 16

DSolve[{y'[x]==-2*y[x]+y[x]^2,{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {2}{e^{2 x}+1} \]