18.8 problem 27.1 (h)

Internal problem ID [13855]
Internal file name [OUTPUT/13027_Friday_February_23_2024_06_46_45_AM_16019750/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 27. Differentiation and the Laplace transform. Additional Exercises. page 496
Problem number: 27.1 (h).
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, _with_linear_symmetries]]

\[ \boxed {y^{\prime \prime }+5 y^{\prime }+6 y={\mathrm e}^{4 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}

18.8.1 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) &=6\\ F &={\mathrm e}^{4 t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+5 y^{\prime }+6 y = {\mathrm e}^{4 t} \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 )+6 Y \left (s \right ) = \frac {1}{s -4}\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 )+6 Y \left (s \right ) = \frac {1}{s -4} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2}+s -19}{\left (s -4\right ) \left (s^{2}+5 s +6\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{42 s -168}-\frac {13}{7 \left (s +3\right )}+\frac {17}{6 \left (s +2\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{42 s -168}\right ) &= \frac {{\mathrm e}^{4 t}}{42}\\ \mathcal {L}^{-1}\left (-\frac {13}{7 \left (s +3\right )}\right ) &= -\frac {13 \,{\mathrm e}^{-3 t}}{7}\\ \mathcal {L}^{-1}\left (\frac {17}{6 \left (s +2\right )}\right ) &= \frac {17 \,{\mathrm e}^{-2 t}}{6} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {13 \,{\mathrm e}^{-3 t}}{7}+\frac {17 \,{\mathrm e}^{-2 t}}{6}+\frac {{\mathrm e}^{4 t}}{42} \] Simplifying the solution gives \[ y = \frac {\left ({\mathrm e}^{7 t}+119 \,{\mathrm e}^{t}-78\right ) {\mathrm e}^{-3 t}}{42} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left ({\mathrm e}^{7 t}+119 \,{\mathrm e}^{t}-78\right ) {\mathrm e}^{-3 t}}{42} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {\left ({\mathrm e}^{7 t}+119 \,{\mathrm e}^{t}-78\right ) {\mathrm e}^{-3 t}}{42} \] Verified OK.

18.8.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+5 y^{\prime }+6 y={\mathrm e}^{4 t}, 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 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+5 r +6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +3\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, -2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-2 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 )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 t} c_{1} +c_{2} {\mathrm e}^{-2 t}+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )={\mathrm e}^{4 t}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{-3 t} & {\mathrm e}^{-2 t} \\ -3 \,{\mathrm e}^{-3 t} & -2 \,{\mathrm e}^{-2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{-5 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-{\mathrm e}^{-3 t} \left (\int {\mathrm e}^{7 t}d t \right )+{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{6 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{4 t}}{42} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-3 t} c_{1} +c_{2} {\mathrm e}^{-2 t}+\frac {{\mathrm e}^{4 t}}{42} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y={\mathrm e}^{-3 t} c_{1} +c_{2} {\mathrm e}^{-2 t}+\frac {{\mathrm e}^{4 t}}{42} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} +\frac {1}{42} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 \,{\mathrm e}^{-3 t} c_{1} -2 c_{2} {\mathrm e}^{-2 t}+\frac {2 \,{\mathrm e}^{4 t}}{21} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-3 c_{1} -2 c_{2} +\frac {2}{21} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {13}{7}, c_{2} =\frac {17}{6}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{7 t}+119 \,{\mathrm e}^{t}-78\right ) {\mathrm e}^{-3 t}}{42} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left ({\mathrm e}^{7 t}+119 \,{\mathrm e}^{t}-78\right ) {\mathrm e}^{-3 t}}{42} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`
 

✓ Solution by Maple

Time used: 4.953 (sec). Leaf size: 23

dsolve([diff(y(t),t$2)+5*diff(y(t),t)+6*y(t)=exp(4*t),y(0) = 1, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left ({\mathrm e}^{7 t}+119 \,{\mathrm e}^{t}-78\right ) {\mathrm e}^{-3 t}}{42} \]

✓ Solution by Mathematica

Time used: 0.048 (sec). Leaf size: 26

DSolve[{y''[t]+5*y'[t]+6*y[t]==Exp[4*t],{y[0]==1,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{42} e^{-3 t} \left (119 e^t+e^{7 t}-78\right ) \]