5.2 problem 32

Internal problem ID [6657]
Internal file name [OUTPUT/5905_Sunday_June_05_2022_04_01_02_PM_89408436/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.2.2 TRANSFORMS OF DERIVATIVES Page 289
Problem number: 32.
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]

\[ \boxed {2 y^{\prime }+y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -3] \end {align*}

5.2.1 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) &={\frac {1}{2}}\\ q(t) &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {y}{2} = 0 \end {align*}

The domain of \(p(t)={\frac {1}{2}}\) is \[ \{-\infty

5.2.2 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*} 2 s Y \left (s \right )-2 y \left (0\right )+Y \left (s \right ) = 0\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} 2 s Y \left (s \right )+6+Y \left (s \right ) = 0 \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {6}{2 s +1} \end {align*}

Taking inverse Laplace transform gives \[ \mathcal {L}^{-1}\left (-\frac {6}{2 s +1}\right ) = -3 \,{\mathrm e}^{-\frac {t}{2}} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -3 \,{\mathrm e}^{-\frac {t}{2}} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -3 \,{\mathrm e}^{-\frac {t}{2}} \] Verified OK.

5.2.3 Maple step by step solution

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

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: 1.438 (sec). Leaf size: 10

dsolve([2*diff(y(t),t)+y(t)=0,y(0) = -3],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.022 (sec). Leaf size: 14

DSolve[{2*y'[t]+y[t]==0,{y[0]==-3}},y[t],t,IncludeSingularSolutions -> True]
 

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