13.8 problem 8

Internal problem ID [12791]
Internal file name [OUTPUT/11444_Saturday_November_04_2023_08_47_21_AM_50415031/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.2, page 248
Problem number: 8.
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]

\[ \boxed {y^{\prime }={\mathrm e}^{x}} \] With initial conditions \begin {align*} [y \left (0\right ) = -1] \end {align*}

13.8.1 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) &=0\\ q(x) &={\mathrm e}^{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = {\mathrm e}^{x} \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

13.8.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right ) = \frac {1}{s -1}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )+1 = \frac {1}{s -1} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {-2+s}{\left (s -1\right ) s} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {2}{s}+\frac {1}{s -1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {2}{s}\right ) &= -2\\ \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{x} \end {align*}

Adding the above results and simplifying gives \[ y=-2+{\mathrm e}^{x} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -2+{\mathrm e}^{x} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -2+{\mathrm e}^{x} \] Verified OK.

13.8.3 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

✓ Solution by Maple

Time used: 5.328 (sec). Leaf size: 8

dsolve([diff(y(x),x)=exp(x),y(0) = -1],y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 10

DSolve[{y'[x]==Exp[x],{y[0]==-1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^x-2 \]