Internal problem ID [866]
Internal file name [OUTPUT/866_Sunday_June_05_2022_01_52_46_AM_63188027/index.tex]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 6.5, The Laplace Transform. Impulse functions. page 273
Problem number: 12.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime }+y=\frac {\operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )}{2 k}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \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) &=0\\ q(t) &=1\\ F &=\frac {\operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )}{2 k} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+y = \frac {\operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )}{2 k} \end {align*}
The domain of \(p(t)=0\) 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*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+Y \left (s \right ) = \frac {\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4+k \right ), t , s\right )}{2 k}-\frac {\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4-k \right ), t , s\right )}{2 k}\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 )+Y \left (s \right ) = \frac {\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4+k \right ), t , s\right )}{2 k}-\frac {\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4-k \right ), t , s\right )}{2 k} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4+k \right ), t , s\right )-\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4-k \right ), t , s\right )}{2 k \left (s^{2}+1\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4+k \right ), t , s\right )-\operatorname {laplace} \left (\operatorname {Heaviside}\left (t -4-k \right ), t , s\right )}{2 k \left (s^{2}+1\right )}\right )\\ &= \frac {\left (-\cos \left (-t +4+k \right )+\cos \left (t \right )\right ) \operatorname {Heaviside}\left (-4-k \right )+\operatorname {Heaviside}\left (-4+k \right ) \left (\cos \left (t -4+k \right )-\cos \left (t \right )\right )+\operatorname {Heaviside}\left (t -4-k \right ) \left (\cos \left (-t +4+k \right )-1\right )+\left (-\cos \left (t -4+k \right )+1\right ) \operatorname {Heaviside}\left (t -4+k \right )}{2 k} \end {align*}
Simplifying the solution gives \[
y = \frac {\left (\operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (t -4-k \right )-1\right ) \cos \left (-t +4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )+\left (-\cos \left (t -4+k \right )+1\right ) \operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t \right )-\cos \left (t \right ) \operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t -4+k \right )+\cos \left (t \right )}{2 k}
\] The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\left (\operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (t -4-k \right )-1\right ) \cos \left (-t +4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )+\left (-\cos \left (t -4+k \right )+1\right ) \operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t \right )-\cos \left (t \right ) \operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t -4+k \right )+\cos \left (t \right )}{2 k} \\
\end{align*} Verification of solutions
\[
y = \frac {\left (\operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (t -4-k \right )-1\right ) \cos \left (-t +4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )+\left (-\cos \left (t -4+k \right )+1\right ) \operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t \right )-\cos \left (t \right ) \operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t -4+k \right )+\cos \left (t \right )}{2 k}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+y=\frac {\mathit {Heaviside}\left (t -4+k \right )-\mathit {Heaviside}\left (t -4-k \right )}{2 k}, 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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (t \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} \cos \left (t \right )+c_{2} \sin \left (t \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 {\mathit {Heaviside}\left (t -4+k \right )-\mathit {Heaviside}\left (t -4-k \right )}{2 k}\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 (t \right ) & \sin \left (t \right ) \\ -\sin \left (t \right ) & \cos \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {-\cos \left (t \right ) \left (\int \sin \left (t \right ) \left (\mathit {Heaviside}\left (t -4+k \right )-\mathit {Heaviside}\left (t -4-k \right )\right )d t \right )+\sin \left (t \right ) \left (\int \cos \left (t \right ) \left (\mathit {Heaviside}\left (t -4+k \right )-\mathit {Heaviside}\left (t -4-k \right )\right )d t \right )}{2 k} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\mathit {Heaviside}\left (t -4-k \right ) \left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-1\right )-\mathit {Heaviside}\left (t -4+k \right ) \left (\cos \left (t \right ) \cos \left (-4+k \right )-\sin \left (t \right ) \sin \left (-4+k \right )-1\right )}{2 k} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right )+\frac {\mathit {Heaviside}\left (t -4-k \right ) \left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-1\right )-\mathit {Heaviside}\left (t -4+k \right ) \left (\cos \left (t \right ) \cos \left (-4+k \right )-\sin \left (t \right ) \sin \left (-4+k \right )-1\right )}{2 k} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right )+\frac {\mathit {Heaviside}\left (t -4-k \right ) \left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-1\right )-\mathit {Heaviside}\left (t -4+k \right ) \left (\cos \left (t \right ) \cos \left (-4+k \right )-\sin \left (t \right ) \sin \left (-4+k \right )-1\right )}{2 k} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +\frac {\mathit {Heaviside}\left (-4-k \right ) \left (\cos \left (4+k \right )-1\right )-\mathit {Heaviside}\left (-4+k \right ) \left (\cos \left (-4+k \right )-1\right )}{2 k} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right )+\frac {\mathit {Dirac}\left (-t +4+k \right ) \left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-1\right )+\mathit {Heaviside}\left (t -4-k \right ) \left (-\sin \left (t \right ) \cos \left (4+k \right )+\cos \left (t \right ) \sin \left (4+k \right )\right )-\mathit {Dirac}\left (t -4+k \right ) \left (\cos \left (t \right ) \cos \left (-4+k \right )-\sin \left (t \right ) \sin \left (-4+k \right )-1\right )-\mathit {Heaviside}\left (t -4+k \right ) \left (-\sin \left (t \right ) \cos \left (-4+k \right )-\cos \left (t \right ) \sin \left (-4+k \right )\right )}{2 k} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{2} +\frac {\mathit {Dirac}\left (4+k \right ) \left (\cos \left (4+k \right )-1\right )+\sin \left (4+k \right ) \mathit {Heaviside}\left (-4-k \right )-\mathit {Dirac}\left (-4+k \right ) \left (\cos \left (-4+k \right )-1\right )+\sin \left (-4+k \right ) \mathit {Heaviside}\left (-4+k \right )}{2 k} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {\mathit {Heaviside}\left (-4+k \right ) \cos \left (-4+k \right )-\cos \left (4+k \right ) \mathit {Heaviside}\left (-4-k \right )-\mathit {Heaviside}\left (-4+k \right )+\mathit {Heaviside}\left (-4-k \right )}{2 k}, c_{2} =-\frac {\sin \left (-4+k \right ) \mathit {Heaviside}\left (-4+k \right )+\sin \left (4+k \right ) \mathit {Heaviside}\left (-4-k \right )+\mathit {Dirac}\left (4+k \right ) \cos \left (4+k \right )-\mathit {Dirac}\left (-4+k \right ) \cos \left (-4+k \right )-\mathit {Dirac}\left (4+k \right )+\mathit {Dirac}\left (-4+k \right )}{2 k}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (t -4-k \right ) \left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-1\right )+\left (-\cos \left (t \right ) \cos \left (-4+k \right )+\sin \left (t \right ) \sin \left (-4+k \right )+1\right ) \mathit {Heaviside}\left (t -4+k \right )+\mathit {Heaviside}\left (-4+k \right ) \left (\cos \left (t \right ) \cos \left (-4+k \right )-\sin \left (t \right ) \sin \left (-4+k \right )-\cos \left (t \right )\right )+\left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-\cos \left (t \right )\right ) \left (\mathit {Heaviside}\left (4+k \right )-1\right )}{2 k} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (t -4-k \right ) \left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-1\right )+\left (-\cos \left (t \right ) \cos \left (-4+k \right )+\sin \left (t \right ) \sin \left (-4+k \right )+1\right ) \mathit {Heaviside}\left (t -4+k \right )+\mathit {Heaviside}\left (-4+k \right ) \left (\cos \left (t \right ) \cos \left (-4+k \right )-\sin \left (t \right ) \sin \left (-4+k \right )-\cos \left (t \right )\right )+\left (\cos \left (t \right ) \cos \left (4+k \right )+\sin \left (t \right ) \sin \left (4+k \right )-\cos \left (t \right )\right ) \left (\mathit {Heaviside}\left (4+k \right )-1\right )}{2 k} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.641 (sec). Leaf size: 76
\[
y \left (t \right ) = \frac {\left (\operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (t -4-k \right )-1\right ) \cos \left (-t +4+k \right )-\operatorname {Heaviside}\left (t -4-k \right )+\left (-\cos \left (t -4+k \right )+1\right ) \operatorname {Heaviside}\left (t -4+k \right )-\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t \right )-\cos \left (t \right ) \operatorname {Heaviside}\left (4+k \right )+\operatorname {Heaviside}\left (-4+k \right ) \cos \left (t -4+k \right )+\cos \left (t \right )}{2 k}
\]
✓ Solution by Mathematica
Time used: 1.204 (sec). Leaf size: 181
\begin{align*}
y(t)\to \fbox {$\frac {(\cos (k-t+4)-1) \theta (-k+t-4)-(\cos (-k-t+4)-1) \theta (k+t-4)}{2 k}\text { if }-45.11.2 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([diff(y(t),t$2)+y(t)=1/(2*k)*(Heaviside(t-(4-k)) - Heaviside(t-(4+k)) ),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+y[t]==1/(2*k)*(UnitStep[t-(4-k)] - UnitStep[t-(4+k)] ),{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]