3.42 problem 4.8 (a)

Internal problem ID [13339]
Internal file name [OUTPUT/12511_Wednesday_February_14_2024_11_54_29_PM_15906252/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 4. SEPARABLE FIRST ORDER EQUATIONS. Additional exercises. page 90
Problem number: 4.8 (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]

\[ \boxed {y^{\prime }-2 y=-10} \] With initial conditions \begin {align*} [y \left (0\right ) = 8] \end {align*}

3.42.1 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) &=-10 \end {align*}

Hence the ode is \begin {align*} y^{\prime }-2 y = -10 \end {align*}

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

3.42.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{2 y -10}d y &= \int {dx}\\ \frac {\ln \left (y -5\right )}{2}&= x +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=8\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\ln \left (3\right )}{2} = c_{1} \end {align*}

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

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

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y -5\right )}{2} = x +\frac {\ln \left (3\right )}{2} \end {align*}

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (y-5\right )}{2} &= x +\frac {\ln \left (3\right )}{2} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (y-5\right )}{2} = x +\frac {\ln \left (3\right )}{2} \] Verified OK.

3.42.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y=-10, y \left (0\right )=8\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 y-10 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{2 y-10}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{2 y-10}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y-5\right )}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{2 x +2 c_{1}}+5 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=8 \\ {} & {} & 8={\mathrm e}^{2 c_{1}}+5 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\ln \left (3\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\ln \left (3\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=3 \,{\mathrm e}^{2 x}+5 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=3 \,{\mathrm e}^{2 x}+5 \end {array} \]

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: 12

dsolve([diff(y(x),x)-2*y(x)=-10,y(0) = 8],y(x), singsol=all)
 

\[ y \left (x \right ) = 5+3 \,{\mathrm e}^{2 x} \]

✓ Solution by Mathematica

Time used: 0.055 (sec). Leaf size: 14

DSolve[{y'[x]-2*y[x]==-10,{y[0]==8}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 3 e^{2 x}+5 \]