7.1 problem 1.2-2 (a)

Internal problem ID [2476]
Internal file name [OUTPUT/1968_Sunday_June_05_2022_02_41_23_AM_85041395/index.tex]

Book: Ordinary Differential Equations, Robert H. Martin, 1983
Section: Problem 1.2-2, page 12
Problem number: 1.2-2 (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]

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

7.1.1 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) &=-1\\ q(t) &=0 \end {align*}

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

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

7.1.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{y}d y &= t +c_{1}\\ \ln \left (y \right )&=t +c_{1}\\ y&={\mathrm e}^{t +c_{1}}\\ y&={\mathrm e}^{t} c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 2 \end {align*}

Trying the constant \begin {align*} c_{1} = 2 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=2 \,{\mathrm e}^{t} \end {align*}

The constant \(c_{1} = 2\) gives valid solution.

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

7.1.3 Maple step by step solution

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

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: 0.015 (sec). Leaf size: 8

dsolve([diff(y(t),t)=y(t),y(0) = 2],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.023 (sec). Leaf size: 10

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

\[ y(t)\to 2 e^t \]