5.7 problem 2 and 14(iii)

Internal problem ID [12956]
Internal file name [OUTPUT/11609_Tuesday_November_07_2023_11_51_59_PM_96989353/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(iii).
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 ) = 6] \end {align*}

5.7.1 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

5.7.2 Solving as quadrature ode

Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 6\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=6 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 6 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 6 \] Verified OK.

5.7.3 Maple step by step solution

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

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: 0.016 (sec). Leaf size: 5

dsolve([diff(y(t),t)=y(t)^2-4*y(t)-12,y(0) = 6],y(t), singsol=all)
 

\[ y \left (t \right ) = 6 \]

✓ Solution by Mathematica

Time used: 0.002 (sec). Leaf size: 6

DSolve[{y'[t]==y[t]^2-4*y[t]-12,{y[0]==6}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 6 \]