4.15 problem 15

Internal problem ID [12650]
Internal file name [OUTPUT/11303_Friday_November_03_2023_06_30_05_AM_72975530/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: 15.
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]

\[ \boxed {y^{\prime }-a y=b} \] With initial conditions \begin {align*} [y \left (c \right ) = d] \end {align*}

4.15.1 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) &=-a\\ q(x) &=b \end {align*}

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

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

4.15.2 Solving as quadrature ode

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

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

The solutions are \begin {align*} c_{1} = \frac {-c a +\ln \left (a d +b \right )}{a} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {-c a +\ln \left (a d +b \right )}{a} \end {align*}

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

The constant \(c_{1} = \frac {-c a +\ln \left (a d +b \right )}{a}\) gives valid solution.

Summary

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

Verification of solutions

\[ \frac {\ln \left (a y+b \right )}{a} = \frac {\ln \left (a d +b \right )+\left (-c +x \right ) a}{a} \] Verified OK.

4.15.3 Maple step by step solution

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

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: 27

dsolve([diff(y(x),x)=a*y(x)+b,y(c) = d],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (a d +b \right ) {\mathrm e}^{-a \left (c -x \right )}-b}{a} \]

✓ Solution by Mathematica

Time used: 0.06 (sec). Leaf size: 39

DSolve[{y'[x]==a*y[x]+b,{y[c]==d}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {e^{-a c} \left (b \left (e^{a x}-e^{a c}\right )+a d e^{a x}\right )}{a} \]