3.19 problem 20

Internal problem ID [946]
Internal file name [OUTPUT/946_Sunday_June_05_2022_01_54_52_AM_16422908/index.tex]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 2, First order equations. separable equations. Section 2.2 Page 52
Problem number: 20.
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*}

3.19.1 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

3.19.2 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.

3.19.3 Maple step by step solution

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

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: 0.016 (sec). Leaf size: 14

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

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

✓ Solution by Mathematica

Time used: 0.009 (sec). Leaf size: 21

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

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