5.6 problem 2 and 14(ii)

Internal problem ID [12955]
Internal file name [OUTPUT/11608_Tuesday_November_07_2023_11_51_58_PM_29876747/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(ii).
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 (1\right ) = 0] \end {align*}

5.6.1 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 \(t=1\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=1\) is \[ \{-\infty

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

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

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

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

Verification of solutions

\[ \frac {\ln \left (y-6\right )}{8}-\frac {\ln \left (y+2\right )}{8} = t -1+\frac {i \pi }{8}+\frac {\ln \left (3\right )}{8} \] Verified OK.

5.6.3 Maple step by step solution

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

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: 0.125 (sec). Leaf size: 26

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

\[ y \left (t \right ) = \frac {6-6 \,{\mathrm e}^{-8+8 t}}{3 \,{\mathrm e}^{-8+8 t}+1} \]

✓ Solution by Mathematica

Time used: 0.018 (sec). Leaf size: 32

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

\[ y(t)\to \frac {6 e^8-6 e^{8 t}}{3 e^{8 t}+e^8} \]