18.1 problem 27.1 (a)

Internal problem ID [13848]
Internal file name [OUTPUT/13020_Friday_February_23_2024_06_46_42_AM_76124009/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 27. Differentiation and the Laplace transform. Additional Exercises. page 496
Problem number: 27.1 (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]

\[ \boxed {y^{\prime }+4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 3] \end {align*}

18.1.1 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) &=4\\ q(t) &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime }+4 y = 0 \end {align*}

The domain of \(p(t)=4\) is \[ \{-\infty

18.1.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )+4 Y \left (s \right ) = 0\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-3+4 Y \left (s \right ) = 0 \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {3}{s +4} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {3}{s +4}\right )\\ &= 3 \,{\mathrm e}^{-4 t} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 3 \,{\mathrm e}^{-4 t} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 3 \,{\mathrm e}^{-4 t} \] Verified OK.

18.1.3 Maple step by step solution

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

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: 3.953 (sec). Leaf size: 10

dsolve([diff(y(t),t)+4*y(t)=0,y(0) = 3],y(t), singsol=all)
 

\[ y \left (t \right ) = 3 \,{\mathrm e}^{-4 t} \]

✓ Solution by Mathematica

Time used: 0.047 (sec). Leaf size: 12

DSolve[{y'[t]+4*y[t]==0,{y[0]==3}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 3 e^{-4 t} \]