2.1 problem 1.1-3 (a)

Internal problem ID [2454]
Internal file name [OUTPUT/1946_Sunday_June_05_2022_02_40_22_AM_70399770/index.tex]

Book: Ordinary Differential Equations, Robert H. Martin, 1983
Section: Problem 1.1-3, page 6
Problem number: 1.1-3 (a).
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]

\[ \boxed {y^{\prime }-2 y=-4} \] With initial conditions \begin {align*} [y \left (0\right ) = 5] \end {align*}

2.1.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}

Where here \begin {align*} p(t) &=-2\\ q(t) &=-4 \end {align*}

Hence the ode is \begin {align*} y^{\prime }-2 y = -4 \end {align*}

The domain of \(p(t)=-2\) is \[ \{-\infty

2.1.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=5\) 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\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\right )}{2} &= t +\frac {\ln \left (3\right )}{2} \\ \end{align*}

Verification of solutions

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

2.1.3 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 12

dsolve([diff(y(t),t)=2*y(t)-4,y(0) = 5],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.025 (sec). Leaf size: 14

DSolve[{y'[t]==2*y[t]-4,y[0]==5},y[t],t,IncludeSingularSolutions -> True]
 

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