problem 3

Internal problem ID [858]
Internal ﬁle name [OUTPUT/858_Sunday_June_05_2022_01_52_10_AM_85375152/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: 3.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+3 y^{\prime }+2 y=\delta \left (t -5\right )+\operatorname {Heaviside}\left (t -10\right )} \] With initial conditions \begin {align*} \left [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = {\frac {1}{2}}\right ] \end {align*}

5.3.1 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) &=3\\ q(t) &=2\\ F &=\delta \left (t -5\right )+\operatorname {Heaviside}\left (t -10\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+3 y^{\prime }+2 y = \delta \left (t -5\right )+\operatorname {Heaviside}\left (t -10\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 )+3 s Y \left (s \right )-3 y \left (0\right )+2 Y \left (s \right ) = {\mathrm e}^{-5 s}+\frac {{\mathrm e}^{-10 s}}{s}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-\frac {1}{2}+3 s Y \left (s \right )+2 Y \left (s \right ) = {\mathrm e}^{-5 s}+\frac {{\mathrm e}^{-10 s}}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 \,{\mathrm e}^{-5 s} s +2 \,{\mathrm e}^{-10 s}+s}{2 s \left (s^{2}+3 s +2\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {2 \,{\mathrm e}^{-5 s} s +2 \,{\mathrm e}^{-10 s}+s}{2 s \left (s^{2}+3 s +2\right )}\right )\\ &= -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}+\frac {\operatorname {Heaviside}\left (t -10\right ) \left (1+{\mathrm e}^{-2 t +20}-2 \,{\mathrm e}^{10-t}\right )}{2}+\operatorname {Heaviside}\left (t -5\right ) \left (-{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5}\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2} & t <5 \\ -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}-{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & t <10 \\ -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}+\frac {1}{2}+\frac {{\mathrm e}^{-2 t +20}}{2}-{\mathrm e}^{10-t}-{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & 10\le t \end {array}\right . \] Simplifying the solution gives \[ y = -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}+\left (\left \{\begin {array}{cc} 0 & t <5 \\ -{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & t <10 \\ \frac {1}{2}+\frac {{\mathrm e}^{-2 t +20}}{2}-{\mathrm e}^{10-t}-{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & 10\le t \end {array}\right .\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}+\left (\left \{\begin {array}{cc} 0 & t <5 \\ -{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & t <10 \\ \frac {1}{2}+\frac {{\mathrm e}^{-2 t +20}}{2}-{\mathrm e}^{10-t}-{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & 10\le t \end {array}\right .\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}+\left (\left \{\begin {array}{cc} 0 & t <5 \\ -{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & t <10 \\ \frac {1}{2}+\frac {{\mathrm e}^{-2 t +20}}{2}-{\mathrm e}^{10-t}-{\mathrm e}^{-2 t +10}+{\mathrm e}^{-t +5} & 10\le t \end {array}\right .\right ) \] Veriﬁed OK.

5.3.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+3 y^{\prime }+2 y=\mathit {Dirac}\left (t -5\right )+\mathit {Heaviside}\left (t -10\right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {1}{2}\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}+3 r +2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +2\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, -1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-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}^{-2 t}+c_{2} {\mathrm e}^{-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 )=\mathit {Dirac}\left (t -5\right )+\mathit {Heaviside}\left (t -10\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}^{-2 t} & {\mathrm e}^{-t} \\ -2 \,{\mathrm e}^{-2 t} & -{\mathrm e}^{-t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-{\mathrm e}^{-2 t} \left (\int \left ({\mathrm e}^{10} \mathit {Dirac}\left (t -5\right )+\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{2 t}\right )d t \right )+{\mathrm e}^{-t} \left (\int \left (\mathit {Dirac}\left (t -5\right )+\mathit {Heaviside}\left (t -10\right )\right ) {\mathrm e}^{t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-2 t +10}+\frac {\mathit {Heaviside}\left (t -10\right )}{2}+\frac {\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{-2 t +20}}{2}-\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t}-\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-2 t +10}+\frac {\mathit {Heaviside}\left (t -10\right )}{2}+\frac {\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{-2 t +20}}{2}-\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t}-\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-2 t +10}+\frac {\mathit {Heaviside}\left (t -10\right )}{2}+\frac {\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{-2 t +20}}{2}-\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}-c_{2} {\mathrm e}^{-t}-\mathit {Dirac}\left (t -5\right ) {\mathrm e}^{-2 t +10}+2 \mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-2 t +10}+\frac {\mathit {Dirac}\left (t -10\right )}{2}+\frac {\mathit {Dirac}\left (t -10\right ) {\mathrm e}^{-2 t +20}}{2}-\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{-2 t +20}-\mathit {Dirac}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Dirac}\left (t -5\right ) {\mathrm e}^{-t +5}-\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {1}{2} \\ {} & {} & \frac {1}{2}=-2 c_{1} -c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{2}, c_{2} =\frac {1}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}-\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-2 t +10}+\frac {\mathit {Heaviside}\left (t -10\right )}{2}+\frac {\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{-2 t +20}}{2}-\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{-t}}{2}-\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-2 t +10}+\frac {\mathit {Heaviside}\left (t -10\right )}{2}+\frac {\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{-2 t +20}}{2}-\mathit {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\mathit {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5} \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)] 
<- double symmetry of the form [xi=0, eta=F(x)] successful`

\[ y \left (t \right ) = \frac {{\mathrm e}^{-t}}{2}-\frac {{\mathrm e}^{-2 t}}{2}-\operatorname {Heaviside}\left (t -10\right ) {\mathrm e}^{10-t}+\frac {\operatorname {Heaviside}\left (t -10\right ) {\mathrm e}^{20-2 t}}{2}+\frac {\operatorname {Heaviside}\left (t -10\right )}{2}+\operatorname {Heaviside}\left (t -5\right ) {\mathrm e}^{-t +5}-\operatorname {Heaviside}\left (t -5\right ) {\mathrm e}^{10-2 t} \]

\[ y(t)\to \frac {1}{2} e^{-2 t} \left (2 e^5 \left (e^t-e^5\right ) \theta (t-5)+\left (e^{10}-e^t\right )^2 (-\theta (10-t))+e^t+e^{2 t}-2 e^{t+10}+e^{20}-1\right ) \]

5.3 problem 3

5.3.1 Existence and uniqueness analysis

5.3.2 Maple step by step solution