5.5 problem 35

Internal problem ID [6660]
Internal file name [OUTPUT/5908_Sunday_June_05_2022_04_01_09_PM_9545444/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: 35.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff"

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

\[ \boxed {y^{\prime \prime }+5 y^{\prime }+4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}

5.5.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=5\\ q(t) &=4\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+5 y^{\prime }+4 y = 0 \end {align*}

The domain of \(p(t)=5\) is \[ \{-\infty

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (y\right ) =Y(s) \end {align*}

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 )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+5 s Y \left (s \right )-5 y \left (0\right )+4 Y \left (s \right ) = 0\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=0 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-5-s +5 s Y \left (s \right )+4 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s +5}{s^{2}+5 s +4} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {4}{3 \left (s +1\right )}-\frac {1}{3 \left (s +4\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {4}{3 \left (s +1\right )}\right ) &= \frac {4 \,{\mathrm e}^{-t}}{3}\\ \mathcal {L}^{-1}\left (-\frac {1}{3 \left (s +4\right )}\right ) &= -\frac {{\mathrm e}^{-4 t}}{3} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \] Simplifying the solution gives \[ y = -\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \] Verified OK.

5.5.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+5 y^{\prime }+4 y=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+5 r +4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +4\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4, -1\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-4 c_{1} {\mathrm e}^{-4 t}-c_{2} {\mathrm e}^{-t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-4 c_{1} -c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{3}, c_{2} =\frac {4}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

✓ Solution by Maple

Time used: 1.688 (sec). Leaf size: 17

dsolve([diff(y(t),t$2)+5*diff(y(t),t)+4*y(t)=0,y(0) = 1, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {{\mathrm e}^{-4 t}}{3}+\frac {4 \,{\mathrm e}^{-t}}{3} \]

✓ Solution by Mathematica

Time used: 0.014 (sec). Leaf size: 23

DSolve[{y''[t]+5*y'[t]+4*y[t]==0,{y[0]==1,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{3} e^{-4 t} \left (4 e^{3 t}-1\right ) \]