13.3 problem Problem 3

Internal problem ID [2841]
Internal file name [OUTPUT/2333_Sunday_June_05_2022_02_59_33_AM_65921000/index.tex]

Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.4. page 689
Problem number: Problem 3.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "linear", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[[_linear, `class A`]]

\[ \boxed {y^{\prime }+3 y=2 \,{\mathrm e}^{-t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 3] \end {align*}

13.3.1 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) &=3\\ q(t) &=2 \,{\mathrm e}^{-t} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+3 y = 2 \,{\mathrm e}^{-t} \end {align*}

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

13.3.2 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 )+3 Y \left (s \right ) = \frac {2}{1+s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-3+3 Y \left (s \right ) = \frac {2}{1+s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {5+3 s}{\left (1+s \right ) \left (s +3\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {2}{s +3}+\frac {1}{1+s} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2}{s +3}\right ) &= 2 \,{\mathrm e}^{-3 t}\\ \mathcal {L}^{-1}\left (\frac {1}{1+s}\right ) &= {\mathrm e}^{-t} \end {align*}

Adding the above results and simplifying gives \[ y=2 \,{\mathrm e}^{-3 t}+{\mathrm e}^{-t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 2 \,{\mathrm e}^{-3 t}+{\mathrm e}^{-t} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 2 \,{\mathrm e}^{-3 t}+{\mathrm e}^{-t} \] Verified OK.

13.3.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+3 y=2 \,{\mathrm e}^{-t}, y \left (0\right )=3\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+2 \,{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+3 y=2 \,{\mathrm e}^{-t} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+3 y\right )=2 \mu \left (t \right ) {\mathrm e}^{-t} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+3 y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=3 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{3 t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int 2 \mu \left (t \right ) {\mathrm e}^{-t}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int 2 \mu \left (t \right ) {\mathrm e}^{-t}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int 2 \mu \left (t \right ) {\mathrm e}^{-t}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{3 t} \\ {} & {} & y=\frac {\int 2 \,{\mathrm e}^{3 t} {\mathrm e}^{-t}d t +c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{2 t}+c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 t} \left ({\mathrm e}^{2 t}+c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=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={\mathrm e}^{-3 t} \left ({\mathrm e}^{2 t}+2\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 t} \left ({\mathrm e}^{2 t}+2\right ) \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: 2.937 (sec). Leaf size: 15

dsolve([diff(y(t),t)+3*y(t)=2*exp(-t),y(0) = 3],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.051 (sec). Leaf size: 18

DSolve[{y'[t]+3*y[t]==2*Exp[-t],{y[0]==3}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^{-3 t} \left (e^{2 t}+2\right ) \]