4.13 problem 13

Internal problem ID [12648]
Internal file name [OUTPUT/11301_Friday_November_03_2023_06_30_04_AM_273934/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: 13.
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]

\[ \boxed {y^{\prime }-4 y=-5} \] With initial conditions \begin {align*} [y \left (1\right ) = 4] \end {align*}

4.13.1 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) &=-4\\ q(x) &=-5 \end {align*}

Hence the ode is \begin {align*} y^{\prime }-4 y = -5 \end {align*}

The domain of \(p(x)=-4\) is \[ \{-\infty

4.13.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{4 y -5}d y &= \int {dx}\\ \frac {\ln \left (y -\frac {5}{4}\right )}{4}&= x +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=4\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2} = 1+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -1+\frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = -1+\frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y -\frac {5}{4}\right )}{4} = x -1+\frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2} \end {align*}

The constant \(c_{1} = -1+\frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\ln \left (2\right )}{2}+\frac {\ln \left (4 y-5\right )}{4} &= x -1+\frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2} \\ \end{align*}

Verification of solutions

\[ -\frac {\ln \left (2\right )}{2}+\frac {\ln \left (4 y-5\right )}{4} = x -1+\frac {\ln \left (11\right )}{4}-\frac {\ln \left (2\right )}{2} \] Verified OK.

4.13.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-4 y=-5, y \left (1\right )=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=4 y-5 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{4 y-5}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{4 y-5}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (4 y-5\right )}{4}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{4 x +4 c_{1}}}{4}+\frac {5}{4} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=4 \\ {} & {} & 4=\frac {{\mathrm e}^{4+4 c_{1}}}{4}+\frac {5}{4} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-1+\frac {\ln \left (11\right )}{4} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-1+\frac {\ln \left (11\right )}{4}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {11 \,{\mathrm e}^{4 x -4}}{4}+\frac {5}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {11 \,{\mathrm e}^{4 x -4}}{4}+\frac {5}{4} \end {array} \]

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: 14

dsolve([diff(y(x),x)=4*y(x)-5,y(1) = 4],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {5}{4}+\frac {11 \,{\mathrm e}^{-4+4 x}}{4} \]

✓ Solution by Mathematica

Time used: 0.043 (sec). Leaf size: 20

DSolve[{y'[x]==4*y[x]-5,{y[1]==4}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {11}{4} e^{4 x-4}+\frac {5}{4} \]