1.40 problem 47

Internal problem ID [14083]
Internal file name [OUTPUT/13764_Saturday_March_02_2024_02_49_19_PM_98207378/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 1. Introduction to Differential Equations. Exercises 1.1, page 10
Problem number: 47.
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]

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

1.40.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(x) &=2\\ q(x) &=0 \end {align*}

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

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

1.40.2 Solving as quadrature ode

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

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

The solutions are \begin {align*} c_{1} = -\frac {\ln \left (2\right )}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = -\frac {\ln \left (2\right )}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {\ln \left (y \right )}{2} = x -\frac {\ln \left (2\right )}{2} \end {align*}

The constant \(c_{1} = -\frac {\ln \left (2\right )}{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\ln \left (y\right )}{2} &= x -\frac {\ln \left (2\right )}{2} \\ \end{align*}

Verification of solutions

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

1.40.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+2 y=0, y \left (0\right )=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-2 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \left (-2\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=c_{1} -2 x \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1} -2 x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\ln \left (2\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\ln \left (2\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-2 x} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{-2 x} \end {array} \]

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: 0.0 (sec). Leaf size: 10

dsolve([diff(y(x),x)+2*y(x)=0,y(0) = 2],y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.072 (sec). Leaf size: 12

DSolve[{y'[x]+2*y[x]==0,{y[0]==2}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 2 e^{-2 x} \]