Internal problem ID [5708]
Internal file name [OUTPUT/4956_Sunday_June_05_2022_03_14_51_PM_21833028/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 6.4, page 230
Problem number: 7.
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, _linear, _nonhomogeneous]]
\[ \boxed {4 y^{\prime \prime }+24 y^{\prime }+37 y=17 \,{\mathrm e}^{-t}+\delta \left (t -\frac {1}{2}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 1] \end {align*}
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) &={\frac {37}{4}}\\ F &=\frac {17 \,{\mathrm e}^{-t}}{4}+\frac {\delta \left (t -\frac {1}{2}\right )}{4} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+6 y^{\prime }+\frac {37 y}{4} = \frac {17 \,{\mathrm e}^{-t}}{4}+\frac {\delta \left (t -\frac {1}{2}\right )}{4} \end {align*}
The domain of \(p(t)=6\) 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*} 4 s^{2} Y \left (s \right )-4 y^{\prime }\left (0\right )-4 s y \left (0\right )+24 s Y \left (s \right )-24 y \left (0\right )+37 Y \left (s \right ) = \frac {17}{s +1}+{\mathrm e}^{-\frac {s}{2}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} 4 s^{2} Y \left (s \right )-28-4 s +24 s Y \left (s \right )+37 Y \left (s \right ) = \frac {17}{s +1}+{\mathrm e}^{-\frac {s}{2}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\frac {s}{2}} s +4 s^{2}+{\mathrm e}^{-\frac {s}{2}}+32 s +45}{\left (s +1\right ) \left (4 s^{2}+24 s +37\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-\frac {s}{2}} s +4 s^{2}+{\mathrm e}^{-\frac {s}{2}}+32 s +45}{\left (s +1\right ) \left (4 s^{2}+24 s +37\right )}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -\frac {1}{2}\right ) {\mathrm e}^{-3 t +\frac {3}{2}} \sin \left (\frac {t}{2}-\frac {1}{4}\right )}{2}+{\mathrm e}^{-t}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} {\mathrm e}^{-t}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right ) & t <\frac {1}{2} \\ {\mathrm e}^{-t}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\frac {{\mathrm e}^{-3 t +\frac {3}{2}} \sin \left (\frac {t}{2}-\frac {1}{4}\right )}{2} & \frac {1}{2}\le t \end {array}\right .
\] Simplifying the solution gives \[
y = {\mathrm e}^{-t}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\left (\left \{\begin {array}{cc} 0 & t <\frac {1}{2} \\ \frac {{\mathrm e}^{-3 t +\frac {3}{2}} \sin \left (\frac {t}{2}-\frac {1}{4}\right )}{2} & \frac {1}{2}\le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= {\mathrm e}^{-t}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\left (\left \{\begin {array}{cc} 0 & t <\frac {1}{2} \\ \frac {{\mathrm e}^{-3 t +\frac {3}{2}} \sin \left (\frac {t}{2}-\frac {1}{4}\right )}{2} & \frac {1}{2}\le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = {\mathrm e}^{-t}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\left (\left \{\begin {array}{cc} 0 & t <\frac {1}{2} \\ \frac {{\mathrm e}^{-3 t +\frac {3}{2}} \sin \left (\frac {t}{2}-\frac {1}{4}\right )}{2} & \frac {1}{2}\le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 y^{\prime \prime }+24 y^{\prime }+37 y=17 \,{\mathrm e}^{-t}+\mathit {Dirac}\left (t -\frac {1}{2}\right ), y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-6 y^{\prime }-\frac {37 y}{4}+\frac {17 \,{\mathrm e}^{-t}}{4}+\frac {\mathit {Dirac}\left (t -\frac {1}{2}\right )}{4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+6 y^{\prime }+\frac {37 y}{4}=\frac {17 \,{\mathrm e}^{-t}}{4}+\frac {\mathit {Dirac}\left (t -\frac {1}{2}\right )}{4} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+6 r +\frac {37}{4}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-6\right )\pm \left (\sqrt {-1}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3-\frac {\mathrm {I}}{2}, -3+\frac {\mathrm {I}}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right ) \\ \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} \cos \left (\frac {t}{2}\right )+c_{2} {\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+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 )=\frac {17 \,{\mathrm e}^{-t}}{4}+\frac {\mathit {Dirac}\left (t -\frac {1}{2}\right )}{4}\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} \cos \left (\frac {t}{2}\right ) & {\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right ) \\ -3 \,{\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right )-\frac {{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )}{2} & -3 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\frac {{\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right )}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {{\mathrm e}^{-6 t}}{2} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{-3 t} \left (\cos \left (\frac {t}{2}\right ) \left (\int \left ({\mathrm e}^{\frac {3}{2}} \sin \left (\frac {1}{4}\right ) \mathit {Dirac}\left (t -\frac {1}{2}\right )+17 \sin \left (\frac {t}{2}\right ) {\mathrm e}^{2 t}\right )d t \right )-\sin \left (\frac {t}{2}\right ) \left (\int \left ({\mathrm e}^{\frac {3}{2}} \cos \left (\frac {1}{4}\right ) \mathit {Dirac}\left (t -\frac {1}{2}\right )+17 \cos \left (\frac {t}{2}\right ) {\mathrm e}^{2 t}\right )d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+2 \,{\mathrm e}^{2 t}\right )}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right )+c_{2} {\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\frac {{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+2 \,{\mathrm e}^{2 t}\right )}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right )+c_{2} {\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\frac {{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+2 {\mathrm e}^{2 t}\right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=1+c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 c_{1} {\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right )-\frac {c_{1} {\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )}{2}-3 c_{2} {\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+\frac {c_{2} {\mathrm e}^{-3 t} \cos \left (\frac {t}{2}\right )}{2}-\frac {3 \,{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+2 \,{\mathrm e}^{2 t}\right )}{2}+\frac {{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Dirac}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+{\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\frac {\cos \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )}{2}+\frac {\sin \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )}{2}\right )+4 \,{\mathrm e}^{2 t}\right )}{2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-3 c_{1} -1+\frac {c_{2}}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =4\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+2 \,{\mathrm e}^{2 t}+8 \sin \left (\frac {t}{2}\right )\right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-3 t} \left ({\mathrm e}^{\frac {3}{2}} \mathit {Heaviside}\left (t -\frac {1}{2}\right ) \left (\sin \left (\frac {t}{2}\right ) \cos \left (\frac {1}{4}\right )-\cos \left (\frac {t}{2}\right ) \sin \left (\frac {1}{4}\right )\right )+2 \,{\mathrm e}^{2 t}+8 \sin \left (\frac {t}{2}\right )\right )}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.281 (sec). Leaf size: 37
\[
y \left (t \right ) = \frac {\operatorname {Heaviside}\left (t -\frac {1}{2}\right ) {\mathrm e}^{-3 t +\frac {3}{2}} \sin \left (-\frac {1}{4}+\frac {t}{2}\right )}{2}+4 \,{\mathrm e}^{-3 t} \sin \left (\frac {t}{2}\right )+{\mathrm e}^{-t}
\]
✓ Solution by Mathematica
Time used: 0.109 (sec). Leaf size: 63
\[
y(t)\to \frac {1}{84} e^{-9 t/2} \left (7 e^{3/4} \left (e^{3 t}-e^{3/2}\right ) \theta (2 t-1)+12 \left (-7 e^{3 t}+17 e^{7 t/2}-3\right )\right )
\]
8.5.2 Maple step by step solution
`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`
dsolve([4*diff(y(t),t$2)+24*diff(y(t),t)+37*y(t)=17*exp(-t)+Dirac(t-1/2),y(0) = 1, D(y)(0) = 1],y(t), singsol=all)
DSolve[{4*y''[t]+24*y'[t]+27*y[t]==17*Exp[-t]+DiracDelta[t-1/2],{y[0]==1,y'[0]==1}},y[t],t,IncludeSingularSolutions -> True]