3.12 problem 8.8

Internal problem ID [11993]
Internal file name [OUTPUT/10646_Saturday_September_02_2023_02_48_57_PM_58898071/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 8, Separable equations. Exercises page 72
Problem number: 8.8.
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]

\[ \boxed {x^{\prime }+x \left (k^{2}+x^{2}\right )=0} \] With initial conditions \begin {align*} [x \left (0\right ) = x_{0}] \end {align*}

3.12.1 Existence and uniqueness analysis

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

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

3.12.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(x=x_{0}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\frac {2 \ln \left (x_{0} \right )-\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -\frac {2 \ln \left (x_{0} \right )-\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = -\frac {2 \ln \left (x_{0} \right )-\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {2 \ln \left (x \right )-\ln \left (k^{2}+x^{2}\right )}{2 k^{2}} = t -\frac {2 \ln \left (x_{0} \right )-\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}} \end {align*}

The constant \(c_{1} = -\frac {2 \ln \left (x_{0} \right )-\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {-2 \ln \left (x\right )+\ln \left (k^{2}+x^{2}\right )}{2 k^{2}} &= t +\frac {-2 \ln \left (x_{0} \right )+\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}} \\ \end{align*}

Verification of solutions

\[ \frac {-2 \ln \left (x\right )+\ln \left (k^{2}+x^{2}\right )}{2 k^{2}} = t +\frac {-2 \ln \left (x_{0} \right )+\ln \left (k^{2}+x_{0}^{2}\right )}{2 k^{2}} \] Verified OK.

3.12.3 Maple step by step solution

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

dsolve([diff(x(t),t)=-x(t)*(k^2+x(t)^2),x(0) = x__0],x(t), singsol=all)
 

\[ \text {No solution found} \]

✓ Solution by Mathematica

Time used: 1.848 (sec). Leaf size: 62

DSolve[{x'[t]==-x[t]*(k^2+x[t]^2),{x[0]==x0}},x[t],t,IncludeSingularSolutions -> True]
 

\begin{align*} x(t)\to -\frac {k}{\sqrt {e^{2 k^2 t} \left (\frac {k^2}{\text {x0}^2}+1\right )-1}} \\ x(t)\to \frac {k}{\sqrt {e^{2 k^2 t} \left (\frac {k^2}{\text {x0}^2}+1\right )-1}} \\ \end{align*}