problem 31.6 (f)

Internal problem ID [13896]
Internal ﬁle name [OUTPUT/13068_Friday_February_23_2024_06_54_30_AM_65630986/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 31. Delta Functions. Additional Exercises. page 572
Problem number: 31.6 (f).
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"

\[ \boxed {y^{\prime \prime }+y=\delta \left (t \right )+\delta \left (t -\pi \right )} \] Since no initial conditions are explicitly given, then let \begin {align*} y \left (0\right )&= c_{1} \\ y'(0) &= c_{2} \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 )+Y \left (s \right ) = 1+{\mathrm e}^{-s \pi }\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} +Y \left (s \right ) = 1+{\mathrm e}^{-s \pi } \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} +{\mathrm e}^{-s \pi }+c_{2} +1}{s^{2}+1} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {s c_{1} +{\mathrm e}^{-s \pi }+c_{2} +1}{s^{2}+1}\right )\\ &= c_{1} \cos \left (t \right )+\sin \left (t \right ) \left (c_{2} +\operatorname {Heaviside}\left (\pi -t \right )\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} c_{1} \cos \left (t \right )+\sin \left (t \right ) \left (c_{2} +1\right ) & t \le \pi \\ c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right ) & \pi

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (t \right )+\sin \left (t \right ) \left (c_{2} +\left (\left \{\begin {array}{cc} 1 & t \le \pi \\ 0 & \pi

Veriﬁcation of solutions

\[ y = c_{1} \cos \left (t \right )+\sin \left (t \right ) \left (c_{2} +\left (\left \{\begin {array}{cc} 1 & t \le \pi \\ 0 & \pi

22.6.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+y=\mathit {Dirac}\left (t \right )+\mathit {Dirac}\left (t -\pi \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 )=\mathit {Dirac}\left (t \right )+\mathit {Dirac}\left (t -\pi \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 (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 )=\sin \left (t \right ) \left (\int \left (\mathit {Dirac}\left (t \right )-\mathit {Dirac}\left (t -\pi \right )\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\sin \left (t \right ) \left (-\mathit {Heaviside}\left (t -\pi \right )+\mathit {Heaviside}\left (t \right )\right ) \\ \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 )+\sin \left (t \right ) \left (-\mathit {Heaviside}\left (t -\pi \right )+\mathit {Heaviside}\left (t \right )\right ) \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`

\[ y \left (t \right ) = \cos \left (t \right ) y \left (0\right )+\sin \left (t \right ) \left (\operatorname {Heaviside}\left (\pi -t \right )+D\left (y \right )\left (0\right )\right ) \]

\[ y(t)\to -\frac {\theta (t-\pi ) \sin \left (\sqrt {2} (\pi -t)\right )}{\sqrt {2}}+\frac {\theta (t) \sin \left (\sqrt {2} t\right )}{\sqrt {2}}-\frac {\cos \left (\sqrt {2} \pi \right ) \sin \left (\sqrt {2} t\right )}{\sqrt {2}}+c_1 \cos \left (\sqrt {2} t\right )+c_2 \sin \left (\sqrt {2} t\right ) \]

22.6 problem 31.6 (f)

22.6.1 Maple step by step solution