4.21 problem 5.3 (a)

Internal problem ID [13366]
Internal file name [OUTPUT/12538_Wednesday_February_14_2024_11_55_30_PM_61179337/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 5. LINEAR FIRST ORDER EQUATIONS. Additional exercises. page 103
Problem number: 5.3 (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]

\[ \boxed {-3 y+y^{\prime }=6} \] With initial conditions \begin {align*} [y \left (0\right ) = 5] \end {align*}

4.21.1 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) &=6 \end {align*}

Hence the ode is \begin {align*} -3 y+y^{\prime } = 6 \end {align*}

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

4.21.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{3 y +6}d y &= \int {dx}\\ \frac {\ln \left (y +2\right )}{3}&= x +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=5\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\ln \left (7\right )}{3} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\ln \left (7\right )}{3} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\ln \left (7\right )}{3} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y +2\right )}{3} = x +\frac {\ln \left (7\right )}{3} \end {align*}

The constant \(c_{1} = \frac {\ln \left (7\right )}{3}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (y+2\right )}{3} &= x +\frac {\ln \left (7\right )}{3} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (y+2\right )}{3} = x +\frac {\ln \left (7\right )}{3} \] Verified OK.

4.21.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [-3 y+y^{\prime }=6, y \left (0\right )=5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=3 y+6 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{3 y+6}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{3 y+6}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y+2\right )}{3}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{3 c_{1} +3 x}-2 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=5 \\ {} & {} & 5={\mathrm e}^{3 c_{1}}-2 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\ln \left (7\right )}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\ln \left (7\right )}{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=7 \,{\mathrm e}^{3 x}-2 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=7 \,{\mathrm e}^{3 x}-2 \end {array} \]

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: 12

dsolve([diff(y(x),x)-3*y(x)=6,y(0) = 5],y(x), singsol=all)
 

\[ y \left (x \right ) = -2+7 \,{\mathrm e}^{3 x} \]

✓ Solution by Mathematica

Time used: 0.033 (sec). Leaf size: 14

DSolve[{y'[x]-3*y[x]==6,{y[0]==5}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 7 e^{3 x}-2 \]