problem 28.9 (c)

Internal problem ID [13871]
Internal ﬁle name [OUTPUT/13043_Friday_February_23_2024_06_54_20_AM_31494725/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 28. The inverse Laplace transform. Additional Exercises. page 509
Problem number: 28.9 (c).
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }-9 y=24 \,{\mathrm e}^{-3 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 6, y^{\prime }\left (0\right ) = 2] \end {align*}

19.10.1 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) &=0\\ q(t) &=-9\\ F &=24 \,{\mathrm e}^{-3 t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-9 y = 24 \,{\mathrm e}^{-3 t} \end {align*}

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 )-9 Y \left (s \right ) = \frac {24}{s +3}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=6\\ y'(0) &=2 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-2-6 s -9 Y \left (s \right ) = \frac {24}{s +3} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {6 s^{2}+20 s +30}{\left (s +3\right ) \left (s^{2}-9\right )} \end {align*}

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

Adding the above results and simplifying gives \[ y=4 \,{\mathrm e}^{3 t}-2 \,{\mathrm e}^{-3 t} \left (2 t -1\right ) \] Simplifying the solution gives \[ y = \left (-4 t +2\right ) {\mathrm e}^{-3 t}+4 \,{\mathrm e}^{3 t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (-4 t +2\right ) {\mathrm e}^{-3 t}+4 \,{\mathrm e}^{3 t} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (-4 t +2\right ) {\mathrm e}^{-3 t}+4 \,{\mathrm e}^{3 t} \] Veriﬁed OK.

19.10.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-9 y=24 \,{\mathrm e}^{-3 t}, y \left (0\right )=6, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-9=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -3\right ) \left (r +3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 3\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}^{3 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=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{3 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 )=24 \,{\mathrm e}^{-3 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}^{3 t} \\ -3 \,{\mathrm e}^{-3 t} & 3 \,{\mathrm e}^{3 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=6 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-4 \,{\mathrm e}^{-3 t} \left (\int 1d t \right )+4 \,{\mathrm e}^{3 t} \left (\int {\mathrm e}^{-6 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {2 \,{\mathrm e}^{-3 t} \left (6 t +1\right )}{3} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{3 t}-\frac {2 \,{\mathrm e}^{-3 t} \left (6 t +1\right )}{3} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{3 t}-\frac {2 {\mathrm e}^{-3 t} \left (6 t +1\right )}{3} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=6 \\ {} & {} & 6=c_{1} +c_{2} -\frac {2}{3} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t}+3 c_{2} {\mathrm e}^{3 t}+2 \,{\mathrm e}^{-3 t} \left (6 t +1\right )-4 \,{\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-3 c_{1} +3 c_{2} -2 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {8}{3}, c_{2} =4\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (-4 t +2\right ) {\mathrm e}^{-3 t}+4 \,{\mathrm e}^{3 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-4 t +2\right ) {\mathrm e}^{-3 t}+4 \,{\mathrm e}^{3 t} \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`

\[ y \left (t \right ) = \left (-4 t +2\right ) {\mathrm e}^{-3 t}+4 \,{\mathrm e}^{3 t} \]

19.10 problem 28.9 (c)

19.10.1 Existence and uniqueness analysis

19.10.2 Maple step by step solution