problem 19

Internal problem ID [5695]
Internal ﬁle name [OUTPUT/4943_Sunday_June_05_2022_03_11_11_PM_55865171/index.tex]

Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number: 19.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+6 y^{\prime }+8 y={\mathrm e}^{-3 t}-{\mathrm e}^{-5 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}

7.2.1 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) &=6\\ q(t) &=8\\ F &={\mathrm e}^{-3 t}-{\mathrm e}^{-5 t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+6 y^{\prime }+8 y = {\mathrm e}^{-3 t}-{\mathrm e}^{-5 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 )+6 s Y \left (s \right )-6 y \left (0\right )+8 Y \left (s \right ) = \frac {2}{\left (s +3\right ) \left (s +5\right )}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+6 s Y \left (s \right )+8 Y \left (s \right ) = \frac {2}{\left (s +3\right ) \left (s +5\right )} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2}{\left (s +3\right ) \left (s +5\right ) \left (s^{2}+6 s +8\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{3 \left (s +5\right )}+\frac {1}{3 s +6}-\frac {1}{s +3}+\frac {1}{s +4} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{3 \left (s +5\right )}\right ) &= -\frac {{\mathrm e}^{-5 t}}{3}\\ \mathcal {L}^{-1}\left (\frac {1}{3 s +6}\right ) &= \frac {{\mathrm e}^{-2 t}}{3}\\ \mathcal {L}^{-1}\left (-\frac {1}{s +3}\right ) &= -{\mathrm e}^{-3 t}\\ \mathcal {L}^{-1}\left (\frac {1}{s +4}\right ) &= {\mathrm e}^{-4 t} \end {align*}

Adding the above results and simplifying gives \[ y=-{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+{\mathrm e}^{-4 t}+\frac {{\mathrm e}^{-2 t}}{3} \] Simplifying the solution gives \[ y = -{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+{\mathrm e}^{-4 t}+\frac {{\mathrm e}^{-2 t}}{3} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+{\mathrm e}^{-4 t}+\frac {{\mathrm e}^{-2 t}}{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+{\mathrm e}^{-4 t}+\frac {{\mathrm e}^{-2 t}}{3} \] Veriﬁed OK.

7.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+6 y^{\prime }+8 y={\mathrm e}^{-3 t}-{\mathrm e}^{-5 t}, y \left (0\right )=0, 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}+6 r +8=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +4\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4, -2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 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=c_{1} {\mathrm e}^{-4 t}+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}^{-3 t}-{\mathrm e}^{-5 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}^{-4 t} & {\mathrm e}^{-2 t} \\ -4 \,{\mathrm e}^{-4 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 )=2 \,{\mathrm e}^{-6 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-4 t} \left (\int \left ({\mathrm e}^{t}-{\mathrm e}^{-t}\right )d t \right )}{2}+\frac {{\mathrm e}^{-2 t} \left (\int \left ({\mathrm e}^{2 t}-1\right ) {\mathrm e}^{-3 t}d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-5 t}}{3}-{\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-2 t}-\frac {{\mathrm e}^{-5 t}}{3}-{\mathrm e}^{-3 t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-4 t}+c_{2} {\mathrm e}^{-2 t}-\frac {{\mathrm e}^{-5 t}}{3}-{\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {4}{3} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-4 c_{1} {\mathrm e}^{-4 t}-2 c_{2} {\mathrm e}^{-2 t}+\frac {5 \,{\mathrm e}^{-5 t}}{3}+3 \,{\mathrm e}^{-3 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} -2 c_{2} +\frac {14}{3} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =\frac {1}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+{\mathrm e}^{-4 t}+\frac {{\mathrm e}^{-2 t}}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+{\mathrm e}^{-4 t}+\frac {{\mathrm e}^{-2 t}}{3} \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)] 
<- double symmetry of the form [xi=0, eta=F(x)] successful`

\[ y \left (t \right ) = {\mathrm e}^{-4 t}-{\mathrm e}^{-3 t}-\frac {{\mathrm e}^{-5 t}}{3}+\frac {{\mathrm e}^{-2 t}}{3} \]

7.2 problem 19

7.2.1 Existence and uniqueness analysis

7.2.2 Maple step by step solution