1.31 problem 34

1.31.1 Existence and uniqueness analysis
1.31.2 Solving as quadrature ode

Internal problem ID [12894]
Internal file name [OUTPUT/11547_Monday_November_06_2023_01_33_14_PM_20960949/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 1.2. page 33
Problem number: 34.
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]

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

1.31.1 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)\\ &= -\frac {y^{2}-1}{y} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \{y <0\boldsymbol {\lor }0

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \{y <0\boldsymbol {\lor }0

1.31.2 Solving as quadrature ode

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

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

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

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {\ln \left (y^{2}-1\right )}{2} = t -\frac {\ln \left (3\right )}{2} \end {align*}

The constant \(c_{1} = -\frac {\ln \left (3\right )}{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\ln \left (y^{2}-1\right )}{2} &= t -\frac {\ln \left (3\right )}{2} \\ \end{align*}

Verification of solutions

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

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: 0.094 (sec). Leaf size: 16

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

\[ y \left (t \right ) = -\sqrt {3 \,{\mathrm e}^{-2 t}+1} \]

Solution by Mathematica

Time used: 0.013 (sec). Leaf size: 20

DSolve[{y'[t]==(1-y[t]^2)/y[t],{y[0]==-2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -\sqrt {3 e^{-2 t}+1} \]