problem Problem 2(i)[j]

Internal problem ID [12317]
Internal ﬁle name [OUTPUT/10970_Monday_October_02_2023_02_47_39_AM_81295945/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 2(i)[j].
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+8 y^{\prime }+20 y=\sin \left (2 t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -4] \end {align*}

4.10.1 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) &=8\\ q(t) &=20\\ F &=\sin \left (2 t \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+8 y^{\prime }+20 y = \sin \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 )+8 s Y \left (s \right )-8 y \left (0\right )+20 Y \left (s \right ) = \frac {2}{s^{2}+4}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=-4 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-4-s +8 s Y \left (s \right )+20 Y \left (s \right ) = \frac {2}{s^{2}+4} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{3}+4 s^{2}+4 s +18}{\left (s^{2}+4\right ) \left (s^{2}+8 s +20\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {1}{64}-\frac {i}{64}}{s -2 i}+\frac {-\frac {1}{64}+\frac {i}{64}}{s +2 i}+\frac {\frac {33}{64}-\frac {i}{64}}{s +4-2 i}+\frac {\frac {33}{64}+\frac {i}{64}}{s +4+2 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {1}{64}-\frac {i}{64}}{s -2 i}\right ) &= \left (-\frac {1}{64}-\frac {i}{64}\right ) {\mathrm e}^{2 i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{64}+\frac {i}{64}}{s +2 i}\right ) &= \left (-\frac {1}{64}+\frac {i}{64}\right ) {\mathrm e}^{-2 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {33}{64}-\frac {i}{64}}{s +4-2 i}\right ) &= \left (\frac {33}{64}-\frac {i}{64}\right ) {\mathrm e}^{\left (-4+2 i\right ) t}\\ \mathcal {L}^{-1}\left (\frac {\frac {33}{64}+\frac {i}{64}}{s +4+2 i}\right ) &= \left (\frac {33}{64}+\frac {i}{64}\right ) {\mathrm e}^{\left (-4-2 i\right ) t} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {\cos \left (2 t \right ) \left (1-33 \,{\mathrm e}^{-4 t}\right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \] Simplifying the solution gives \[ y = \frac {\left (-1+33 \,{\mathrm e}^{-4 t}\right ) \cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-1+33 \,{\mathrm e}^{-4 t}\right ) \cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {\left (-1+33 \,{\mathrm e}^{-4 t}\right ) \cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \] Veriﬁed OK.

4.10.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+8 y^{\prime }+20 y=\sin \left (2 t \right ), y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4\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}+8 r +20=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-8\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4-2 \,\mathrm {I}, -4+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (2 t \right ) {\mathrm e}^{-4 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (2 t \right ) {\mathrm e}^{-4 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} \cos \left (2 t \right ) {\mathrm e}^{-4 t}+c_{2} \sin \left (2 t \right ) {\mathrm e}^{-4 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 )=\sin \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} \cos \left (2 t \right ) {\mathrm e}^{-4 t} & \sin \left (2 t \right ) {\mathrm e}^{-4 t} \\ -2 \sin \left (2 t \right ) {\mathrm e}^{-4 t}-4 \cos \left (2 t \right ) {\mathrm e}^{-4 t} & 2 \cos \left (2 t \right ) {\mathrm e}^{-4 t}-4 \sin \left (2 t \right ) {\mathrm e}^{-4 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}^{-8 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 (\sin \left (2 t \right ) \left (\int \sin \left (4 t \right ) {\mathrm e}^{4 t}d t \right )-2 \cos \left (2 t \right ) \left (\int {\mathrm e}^{4 t} \sin \left (2 t \right )^{2}d t \right )\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right )}{32} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (2 t \right ) {\mathrm e}^{-4 t}+c_{2} \sin \left (2 t \right ) {\mathrm e}^{-4 t}-\frac {\cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right )}{32} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (2 t \right ) {\mathrm e}^{-4 t}+c_{2} \sin \left (2 t \right ) {\mathrm e}^{-4 t}-\frac {\cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right )}{32} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} -\frac {1}{32} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} \sin \left (2 t \right ) {\mathrm e}^{-4 t}-4 c_{1} \cos \left (2 t \right ) {\mathrm e}^{-4 t}+2 c_{2} \cos \left (2 t \right ) {\mathrm e}^{-4 t}-4 c_{2} \sin \left (2 t \right ) {\mathrm e}^{-4 t}+\frac {\sin \left (2 t \right )}{16}+\frac {\cos \left (2 t \right )}{16} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-4 \\ {} & {} & -4=\frac {1}{16}-4 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {33}{32}, c_{2} =\frac {1}{32}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (-1+33 \,{\mathrm e}^{-4 t}\right ) \cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (-1+33 \,{\mathrm e}^{-4 t}\right ) \cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \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`

\[ y \left (t \right ) = \frac {\left (-1+33 \,{\mathrm e}^{-4 t}\right ) \cos \left (2 t \right )}{32}+\frac {\sin \left (2 t \right ) \left (1+{\mathrm e}^{-4 t}\right )}{32} \]

\[ y(t)\to \frac {1}{32} e^{-4 t} \left (\left (e^{4 t}+1\right ) \sin (2 t)-\left (e^{4 t}-33\right ) \cos (2 t)\right ) \]

4.10 problem Problem 2(i)[j]

4.10.1 Existence and uniqueness analysis

4.10.2 Maple step by step solution