problem Problem 5(e)

Internal problem ID [12341]
Internal ﬁle name [OUTPUT/10994_Monday_October_02_2023_02_47_46_AM_73489955/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 5(e).
ODE order: 2.
ODE degree: 1.

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

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

4.34.1 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) &=1\\ q(t) &={\frac {1}{4}}\\ F &=\frac {{\mathrm e}^{-\frac {1}{2}} \delta \left (t -1\right )}{4} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime }+\frac {y}{4} = \frac {{\mathrm e}^{-\frac {1}{2}} \delta \left (t -1\right )}{4} \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*} 4 s^{2} Y \left (s \right )-4 y^{\prime }\left (0\right )-4 s y \left (0\right )+4 s Y \left (s \right )-4 y \left (0\right )+Y \left (s \right ) = {\mathrm e}^{-\frac {1}{2}-s}\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*} 4 s^{2} Y \left (s \right )+4 s Y \left (s \right )+Y \left (s \right ) = {\mathrm e}^{-\frac {1}{2}-s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {{\mathrm e}^{-\frac {1}{2}-s}}{4 s^{2}+4 s +1} \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 {1}{2}-s}}{4 s^{2}+4 s +1}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} 0 & t <1 \\ \frac {\left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} & 1\le t \end {array}\right . \] Simplifying the solution gives \[ y = \left \{\begin {array}{cc} 0 & t <1 \\ \frac {\left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} & 1\le t \end {array}\right . \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} 0 & t <1 \\ \frac {\left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} & 1\le t \end {array}\right . \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left \{\begin {array}{cc} 0 & t <1 \\ \frac {\left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} & 1\le t \end {array}\right . \] Veriﬁed OK.

4.34.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 y^{\prime \prime }+4 y^{\prime }+y={\mathrm e}^{-\frac {1}{2}} \mathit {Dirac}\left (t -1\right ), 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-y^{\prime }-\frac {y}{4}+\frac {{\mathrm e}^{-\frac {1}{2}} \mathit {Dirac}\left (t -1\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 }+y^{\prime }+\frac {y}{4}=\frac {{\mathrm e}^{-\frac {1}{2}} \mathit {Dirac}\left (t -1\right )}{4} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+r +\frac {1}{4}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (2 r +1\right )^{2}}{4}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =-\frac {1}{2} \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-\frac {t}{2}} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \,{\mathrm e}^{-\frac {t}{2}} \\ \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}^{-\frac {t}{2}}+c_{2} t \,{\mathrm e}^{-\frac {t}{2}}+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 {{\mathrm e}^{-\frac {1}{2}} \mathit {Dirac}\left (t -1\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}^{-\frac {t}{2}} & t \,{\mathrm e}^{-\frac {t}{2}} \\ -\frac {{\mathrm e}^{-\frac {t}{2}}}{2} & {\mathrm e}^{-\frac {t}{2}}-\frac {t \,{\mathrm e}^{-\frac {t}{2}}}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{-t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{-\frac {t}{2}} \left (\int \mathit {Dirac}\left (t -1\right )d t \right ) \left (t -1\right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {\mathit {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-\frac {t}{2}}+c_{2} t \,{\mathrm e}^{-\frac {t}{2}}+\frac {\mathit {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-\frac {t}{2}}+c_{2} t {\mathrm e}^{-\frac {t}{2}}+\frac {\mathit {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} {\mathrm e}^{-\frac {t}{2}}}{2}+c_{2} {\mathrm e}^{-\frac {t}{2}}-\frac {c_{2} t \,{\mathrm e}^{-\frac {t}{2}}}{2}+\frac {\mathit {Dirac}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4}+\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4}-\frac {\mathit {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{8} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {c_{1}}{2}+c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \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`

\[ y \left (t \right ) = \frac {\operatorname {Heaviside}\left (t -1\right ) \left (t -1\right ) {\mathrm e}^{-\frac {t}{2}}}{4} \]

4.34 problem Problem 5(e)

4.34.1 Existence and uniqueness analysis

4.34.2 Maple step by step solution