Internal problem ID [12348]
Internal file name [OUTPUT/11001_Monday_October_02_2023_02_47_48_AM_24196050/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 14(a).
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
Maple gives the following as the ode type
[[_high_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }-5 y^{\prime \prime }+4 y=12 \operatorname {Heaviside}\left (t \right )-12 \operatorname {Heaviside}\left (t -1\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0] \end {align*}
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] 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 )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime \prime }\right ) &= s^4 Y(s) - y'''(0) - s y''(0) - s^2 y'(0)- s^3 y \left (0\right ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )-5 s^{2} Y \left (s \right )+5 y^{\prime }\left (0\right )+5 s y \left (0\right )+4 Y \left (s \right ) = \frac {12-12 \,{\mathrm e}^{-s}}{s}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=0\\ y^{\prime \prime \prime }\left (0\right )&=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-5 s^{2} Y \left (s \right )+4 Y \left (s \right ) = \frac {12-12 \,{\mathrm e}^{-s}}{s} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = -\frac {12 \left (-1+{\mathrm e}^{-s}\right )}{s \left (s^{4}-5 s^{2}+4\right )} \] Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {12 \left (-1+{\mathrm e}^{-s}\right )}{s \left (s^{4}-5 s^{2}+4\right )}\right )\\ &= -4 \cosh \left (t \right )+\cosh \left (2 t \right )+2 \,{\mathrm e}^{t -1}-\frac {{\mathrm e}^{2 t -2}}{2}+\frac {\operatorname {Heaviside}\left (-t +1\right ) \left (6+{\mathrm e}^{2 t -2}-4 \,{\mathrm e}^{t -1}\right )}{2}+\frac {\operatorname {Heaviside}\left (t -1\right ) \left (4 \,{\mathrm e}^{-t +1}-{\mathrm e}^{-2 t +2}\right )}{2} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -4 \cosh \left (t \right )+\cosh \left (2 t \right )+3 & t <1 \\ -4 \cosh \left (1\right )+\cosh \left (2\right )+\frac {9}{2} & t =1 \\ -4 \cosh \left (t \right )+\cosh \left (2 t \right )+2 \,{\mathrm e}^{t -1}-\frac {{\mathrm e}^{2 t -2}}{2}+2 \,{\mathrm e}^{-t +1}-\frac {{\mathrm e}^{-2 t +2}}{2} & 1 The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} -4 \cosh \left (t \right )+\cosh \left (2 t \right )+3 & t <1 \\ -4 \cosh \left (1\right )+\cosh \left (2\right )+\frac {9}{2} & t &=1 \\ -4 \cosh \left (t \right )+\cosh \left (2 t \right )+2 \,{\mathrm e}^{t -1}-\frac {{\mathrm e}^{2 t -2}}{2}+2 \,{\mathrm e}^{-t +1}-\frac {{\mathrm e}^{-2 t +2}}{2} & 1 Verification of solutions
\[
y = \left \{\begin {array}{cc} -4 \cosh \left (t \right )+\cosh \left (2 t \right )+3 & t <1 \\ -4 \cosh \left (1\right )+\cosh \left (2\right )+\frac {9}{2} & t =1 \\ -4 \cosh \left (t \right )+\cosh \left (2 t \right )+2 \,{\mathrm e}^{t -1}-\frac {{\mathrm e}^{2 t -2}}{2}+2 \,{\mathrm e}^{-t +1}-\frac {{\mathrm e}^{-2 t +2}}{2} & 1 Maple trace
✓ Solution by Maple
Time used: 5.031 (sec). Leaf size: 72
\[
y \left (t \right ) = 2 \left (-1+\cosh \left (t \right )\right )^{2}-\frac {\operatorname {Heaviside}\left (t -1\right ) \left ({\mathrm e}^{-2 t +2}-4 \,{\mathrm e}^{-t +1}+{\mathrm e}^{2 t -2}+6-4 \,{\mathrm e}^{t -1}\right )}{2}
\]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 88
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{2} e^{-2 t} \left (-1+e^t\right )^4 & 0\leq t\leq 1 \\ \frac {1}{2} (-1+e) e^{-2 (t+1)} \left (-e^2-e^3+e^{4 t}+4 e^{t+2}-4 e^{3 t+1}+e^{4 t+1}\right ) & t>1 \\ \end {array} \\ \end {array}
\]
`Methods for high order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 4; linear nonhomogeneous with symmetry [0,1]
trying high order linear exact nonhomogeneous
trying differential order: 4; missing the dependent variable
checking if the LODE has constant coefficients
<- constant coefficients successful`
dsolve([diff(y(t),t$4)-5*diff(y(t),t$2)+4*y(t)=12*(Heaviside(t)-Heaviside(t-1)),y(0) = 0, D(y)(0) = 0, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 0],y(t), singsol=all)
DSolve[{y''''[t]-5*y''[t]+4*y[t]==12*(UnitStep[t]-UnitStep[t-1]),{y[0]==0,y'[0]==0,y''[0]==0,y'''[0]==0}},y[t],t,IncludeSingularSolutions -> True]