5.5 problem 2 and 14(i)

Internal problem ID [12954]
Internal file name [OUTPUT/11607_Tuesday_November_07_2023_11_51_57_PM_65364348/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 1.6 page 89
Problem number: 2 and 14(i).
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]

\[ \boxed {y^{\prime }-y^{2}+4 y=-12} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

5.5.1 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}-4 y -12 \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

5.5.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}-4 y -12}d y &= \int {dt}\\ \frac {\ln \left (y -6\right )}{8}-\frac {\ln \left (y +2\right )}{8}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y -6\right )}{8}-\frac {\ln \left (y +2\right )}{8} = t +\frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8} \end {align*}

The constant \(c_{1} = \frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (y-6\right )}{8}-\frac {\ln \left (y+2\right )}{8} &= t +\frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (y-6\right )}{8}-\frac {\ln \left (y+2\right )}{8} = t +\frac {\ln \left (5\right )}{8}+\frac {i \pi }{8}-\frac {\ln \left (3\right )}{8} \] Verified OK.

5.5.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}+4 y=-12, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-4 y-12 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}-4 y-12}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{2}-4 y-12}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y-6\right )}{8}-\frac {\ln \left (y+2\right )}{8}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {2 \left (3+{\mathrm e}^{8 t +8 c_{1}}\right )}{{\mathrm e}^{8 t +8 c_{1}}-1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {2 \left (3+{\mathrm e}^{8 c_{1}}\right )}{{\mathrm e}^{8 c_{1}}-1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\mathrm {I} \pi }{8}+\frac {\ln \left (\frac {5}{3}\right )}{8} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\mathrm {I} \pi }{8}+\frac {\ln \left (\frac {5}{3}\right )}{8}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {18-10 \,{\mathrm e}^{8 t}}{5 \,{\mathrm e}^{8 t}+3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {18-10 \,{\mathrm e}^{8 t}}{5 \,{\mathrm e}^{8 t}+3} \end {array} \]

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: 0.094 (sec). Leaf size: 23

dsolve([diff(y(t),t)=y(t)^2-4*y(t)-12,y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {18-10 \,{\mathrm e}^{8 t}}{5 \,{\mathrm e}^{8 t}+3} \]

✓ Solution by Mathematica

Time used: 0.018 (sec). Leaf size: 26

DSolve[{y'[t]==y[t]^2-4*y[t]-12,{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {18-10 e^{8 t}}{5 e^{8 t}+3} \]