5.1 problem 31

Internal problem ID [6656]
Internal file name [OUTPUT/5904_Sunday_June_05_2022_04_00_59_PM_36707830/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.2.2 TRANSFORMS OF DERIVATIVES Page 289
Problem number: 31.
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]

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

5.1.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}

Where here \begin {align*} p(t) &=-1\\ q(t) &=1 \end {align*}

Hence the ode is \begin {align*} y^{\prime }-y = 1 \end {align*}

The domain of \(p(t)=-1\) is \[ \{-\infty

5.1.2 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 )-Y \left (s \right ) = \frac {1}{s}\tag {1} \end {align*}

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

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

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

Adding the above results and simplifying gives \[ y={\mathrm e}^{t}-1 \]

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = {\mathrm e}^{t}-1 \] Verified OK.

5.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y=1, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y+1}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y+1\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{t +c_{1}}-1 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0={\mathrm e}^{c_{1}}-1 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{t}-1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{t}-1 \end {array} \]

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: 1.672 (sec). Leaf size: 8

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

\[ y \left (t \right ) = -1+{\mathrm e}^{t} \]

✓ Solution by Mathematica

Time used: 0.029 (sec). Leaf size: 10

DSolve[{y'[t]-y[t]==1,{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^t-1 \]