problem 6

Internal problem ID [5684]
Internal ﬁle name [OUTPUT/4932_Sunday_June_05_2022_03_10_56_PM_44958753/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 6.2, page 216
Problem number: 6.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }-6 y^{\prime }+5 y=29 \cos \left (2 t \right )} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {16}{5}}, y^{\prime }\left (0\right ) = {\frac {31}{5}}\right ] \end {align*}

6.6.1 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) &=5\\ F &=29 \cos \left (2 t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-6 y^{\prime }+5 y = 29 \cos \left (2 t \right ) \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 )+5 Y \left (s \right ) = \frac {29 s}{s^{2}+4}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&={\frac {16}{5}}\\ y'(0) &={\frac {31}{5}} \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+13-\frac {16 s}{5}-6 s Y \left (s \right )+5 Y \left (s \right ) = \frac {29 s}{s^{2}+4} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {16 s^{3}-65 s^{2}+209 s -260}{5 \left (s^{2}+4\right ) \left (s^{2}-6 s +5\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {\frac {1}{10}+\frac {6 i}{5}}{s -2 i}+\frac {\frac {1}{10}-\frac {6 i}{5}}{s +2 i}+\frac {1}{s -1}+\frac {2}{s -5} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\frac {1}{10}+\frac {6 i}{5}}{s -2 i}\right ) &= \left (\frac {1}{10}+\frac {6 i}{5}\right ) {\mathrm e}^{2 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {1}{10}-\frac {6 i}{5}}{s +2 i}\right ) &= \left (\frac {1}{10}-\frac {6 i}{5}\right ) {\mathrm e}^{-2 i t}\\ \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{t}\\ \mathcal {L}^{-1}\left (\frac {2}{s -5}\right ) &= 2 \,{\mathrm e}^{5 t} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+2 \,{\mathrm e}^{5 t}+{\mathrm e}^{t} \] Simplifying the solution gives \[ y = \frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+2 \,{\mathrm e}^{5 t}+{\mathrm e}^{t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+2 \,{\mathrm e}^{5 t}+{\mathrm e}^{t} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+2 \,{\mathrm e}^{5 t}+{\mathrm e}^{t} \] Veriﬁed OK.

6.6.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-6 y^{\prime }+5 y=29 \cos \left (2 t \right ), y \left (0\right )=\frac {16}{5}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {31}{5}\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}-6 r +5=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r -5\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1, 5\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{5 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}^{t}+c_{2} {\mathrm e}^{5 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 )=29 \cos \left (2 t \right )\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}^{t} & {\mathrm e}^{5 t} \\ {\mathrm e}^{t} & 5 \,{\mathrm e}^{5 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=4 \,{\mathrm e}^{6 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {29 \,{\mathrm e}^{t} \left (\int \cos \left (2 t \right ) {\mathrm e}^{-t}d t \right )}{4}+\frac {29 \,{\mathrm e}^{5 t} \left (\int \cos \left (2 t \right ) {\mathrm e}^{-5 t}d t \right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{5 t}+\frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{5 t}+\frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {16}{5} \\ {} & {} & \frac {16}{5}=c_{1} +c_{2} +\frac {1}{5} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{t}+5 c_{2} {\mathrm e}^{5 t}-\frac {2 \sin \left (2 t \right )}{5}-\frac {24 \cos \left (2 t \right )}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {31}{5} \\ {} & {} & \frac {31}{5}=c_{1} +5 c_{2} -\frac {24}{5} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+2 \,{\mathrm e}^{5 t}+{\mathrm e}^{t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+2 \,{\mathrm e}^{5 t}+{\mathrm e}^{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`

\[ y \left (t \right ) = \frac {\cos \left (2 t \right )}{5}-\frac {12 \sin \left (2 t \right )}{5}+{\mathrm e}^{t}+2 \,{\mathrm e}^{5 t} \]

6.6 problem 6

6.6.1 Existence and uniqueness analysis

6.6.2 Maple step by step solution