13.4 problem Problem 4

Internal problem ID [2842]
Internal file name [OUTPUT/2334_Sunday_June_05_2022_02_59_35_AM_75677719/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 4.
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`]]

\[ \boxed {y^{\prime }+2 y=4 t} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

13.4.1 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) &=2\\ q(t) &=4 t \end {align*}

Hence the ode is \begin {align*} y^{\prime }+2 y = 4 t \end {align*}

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

13.4.2 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 )+2 Y \left (s \right ) = \frac {4}{s^{2}}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-1+2 Y \left (s \right ) = \frac {4}{s^{2}} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {s^{2}+4}{s^{2} \left (s +2\right )} \end {align*}

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

Adding the above results and simplifying gives \[ y=2 t +2 \,{\mathrm e}^{-2 t}-1 \]

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 2 t +2 \,{\mathrm e}^{-2 t}-1 \] Verified OK.

13.4.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+2 y=4 t , y \left (0\right )=1\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+4 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 }+2 y=4 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 }+2 y\right )=4 \mu \left (t \right ) 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 }+2 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 )=2 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{2 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 4 \mu \left (t \right ) t d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int 4 \mu \left (t \right ) t d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int 4 \mu \left (t \right ) t d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{2 t} \\ {} & {} & y=\frac {\int 4 \,{\mathrm e}^{2 t} t d t +c_{1}}{{\mathrm e}^{2 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 t -1\right ) {\mathrm e}^{2 t}+c_{1}}{{\mathrm e}^{2 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=2 t -1+c_{1} {\mathrm e}^{-2 t} \\ \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 t +2 \,{\mathrm e}^{-2 t}-1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 t +2 \,{\mathrm e}^{-2 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: 1.703 (sec). Leaf size: 15

dsolve([diff(y(t),t)+2*y(t)=4*t,y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = 2 t +2 \,{\mathrm e}^{-2 t}-1 \]

✓ Solution by Mathematica

Time used: 0.027 (sec). Leaf size: 17

DSolve[{y'[t]+2*y[t]==4*t,{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to 2 t+2 e^{-2 t}-1 \]