4.14 problem 14

Internal problem ID [12649]
Internal file name [OUTPUT/11302_Friday_November_03_2023_06_30_05_AM_62961360/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.2, page 53
Problem number: 14.
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]

\[ \boxed {y^{\prime }+3 y=1} \] With initial conditions \begin {align*} [y \left (-2\right ) = 1] \end {align*}

4.14.1 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) &=3\\ q(x) &=1 \end {align*}

Hence the ode is \begin {align*} y^{\prime }+3 y = 1 \end {align*}

The domain of \(p(x)=3\) is \[ \{-\infty

4.14.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{-3 y +1}d y &= \int {dx}\\ -\frac {\ln \left (y -\frac {1}{3}\right )}{3}&= x +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=-2\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3} = c_{1} -2 \end {align*}

The solutions are \begin {align*} c_{1} = 2-\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3} \end {align*}

Trying the constant \begin {align*} c_{1} = 2-\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {\ln \left (y -\frac {1}{3}\right )}{3} = x +2-\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3} \end {align*}

The constant \(c_{1} = 2-\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (3\right )}{3}-\frac {\ln \left (3 y-1\right )}{3} &= x +2-\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (3\right )}{3}-\frac {\ln \left (3 y-1\right )}{3} = x +2-\frac {\ln \left (2\right )}{3}+\frac {\ln \left (3\right )}{3} \] Verified OK.

4.14.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+3 y=1, y \left (-2\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 y+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-3 y+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-3 y+1}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (-3 y+1\right )}{3}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {{\mathrm e}^{-3 x -3 c_{1}}}{3}+\frac {1}{3} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-2\right )=1 \\ {} & {} & 1=-\frac {{\mathrm e}^{6-3 c_{1}}}{3}+\frac {1}{3} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =2-\frac {\ln \left (2\right )}{3}-\frac {\mathrm {I} \pi }{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =2-\frac {\ln \left (2\right )}{3}-\frac {\mathrm {I} \pi }{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {2 \,{\mathrm e}^{-3 x -6}}{3}+\frac {1}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2 \,{\mathrm e}^{-3 x -6}}{3}+\frac {1}{3} \end {array} \]

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: 14

dsolve([diff(y(x),x)+3*y(x)=1,y(-2) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {1}{3}+\frac {2 \,{\mathrm e}^{-6-3 x}}{3} \]

✓ Solution by Mathematica

Time used: 0.04 (sec). Leaf size: 20

DSolve[{y'[x]+3*y[x]==1,{y[-2]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {2}{3} e^{-3 (x+2)}+\frac {1}{3} \]