Internal problem ID [11521]
Internal file name [OUTPUT/10504_Thursday_May_18_2023_04_21_21_AM_49106921/index.tex]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 3, Laplace transform. Section 3.3 The convolution property. Exercises page
162
Problem number: 7.
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]]
\[ \boxed {x^{\prime \prime }-4 x=1-\operatorname {Heaviside}\left (t -1\right )} \] With initial conditions \begin {align*} [x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 0] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end {align*}
Where here \begin {align*} p(t) &=0\\ q(t) &=-4\\ F &=1-\operatorname {Heaviside}\left (t -1\right ) \end {align*}
Hence the ode is \begin {align*} x^{\prime \prime }-4 x = 1-\operatorname {Heaviside}\left (t -1\right ) \end {align*}
The domain of \(p(t)=0\) is \[
\{-\infty Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (x\right ) =Y(s) \end {align*}
Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (x^{\prime }\right ) &= s Y(s) - x \left (0\right )\\ \mathcal {L}\left (x^{\prime \prime }\right ) &= s^2 Y(s) - x'(0) - s x \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 )-x^{\prime }\left (0\right )-s x \left (0\right )-4 Y \left (s \right ) = \frac {-{\mathrm e}^{-s}+1}{s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} x \left (0\right )&=0\\ x'(0) &=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-4 Y \left (s \right ) = \frac {-{\mathrm e}^{-s}+1}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {{\mathrm e}^{-s}-1}{s \left (s^{2}-4\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} x&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {{\mathrm e}^{-s}-1}{s \left (s^{2}-4\right )}\right )\\ &= -\frac {\operatorname {Heaviside}\left (t -1\right ) \sinh \left (t -1\right )^{2}}{2}-\frac {1}{4}+\frac {\cosh \left (2 t \right )}{4} \end {align*}
Converting the above solution to piecewise it becomes \[
x = \left \{\begin {array}{cc} -\frac {1}{4}+\frac {\cosh \left (2 t \right )}{4} & t <1 \\ -\frac {1}{4}+\frac {\cosh \left (2 t \right )}{4}-\frac {\sinh \left (t -1\right )^{2}}{2} & 1\le t \end {array}\right .
\] Simplifying the solution gives \[
x = -\frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cosh \left (-2+2 t \right ) & 1\le t \end {array}\right .\right )}{4}+\frac {\cosh \left (2 t \right )}{4}
\]
The solution(s) found are the following \begin{align*}
\tag{1} x &= -\frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cosh \left (-2+2 t \right ) & 1\le t \end {array}\right .\right )}{4}+\frac {\cosh \left (2 t \right )}{4} \\
\end{align*} Verification of solutions
\[
x = -\frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cosh \left (-2+2 t \right ) & 1\le t \end {array}\right .\right )}{4}+\frac {\cosh \left (2 t \right )}{4}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }-4 x=1-\mathit {Heaviside}\left (t -1\right ), x \left (0\right )=0, x^{\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 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -2\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{2 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right )+x_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+x_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} x_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} x_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [x_{p}\left (t \right )=-x_{1}\left (t \right ) \left (\int \frac {x_{2}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right )+x_{2}\left (t \right ) \left (\int \frac {x_{1}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=1-\mathit {Heaviside}\left (t -1\right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} \\ -2 \,{\mathrm e}^{-2 t} & 2 \,{\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=4 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=\frac {{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{2 t} \left (-1+\mathit {Heaviside}\left (t -1\right )\right )d t \right )}{4}-\frac {{\mathrm e}^{2 t} \left (\int {\mathrm e}^{-2 t} \left (-1+\mathit {Heaviside}\left (t -1\right )\right )d t \right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\frac {\mathit {Heaviside}\left (t -1\right )}{4}-\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2-2 t}}{8}-\frac {1}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Heaviside}\left (t -1\right )}{8} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+\frac {\mathit {Heaviside}\left (t -1\right )}{4}-\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2-2 t}}{8}-\frac {1}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Heaviside}\left (t -1\right )}{8} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+\frac {\mathit {Heaviside}\left (t -1\right )}{4}-\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2-2 t}}{8}-\frac {1}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Heaviside}\left (t -1\right )}{8} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} -\frac {1}{4} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}+2 c_{2} {\mathrm e}^{2 t}+\frac {\mathit {Dirac}\left (t -1\right )}{4}-\frac {\mathit {Dirac}\left (t -1\right ) {\mathrm e}^{2-2 t}}{8}+\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2-2 t}}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Heaviside}\left (t -1\right )}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Dirac}\left (t -1\right )}{8} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-2 c_{1} +2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {1}{8}, c_{2} =\frac {1}{8}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\mathit {Heaviside}\left (t -1\right )}{4}-\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2-2 t}}{8}-\frac {1}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Heaviside}\left (t -1\right )}{8} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\mathit {Heaviside}\left (t -1\right )}{4}-\frac {\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{2-2 t}}{8}-\frac {1}{4}-\frac {{\mathrm e}^{-2+2 t} \mathit {Heaviside}\left (t -1\right )}{8} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.437 (sec). Leaf size: 24
\[
x \left (t \right ) = -\frac {1}{4}+\frac {\cosh \left (2 t \right )}{4}-\frac {\operatorname {Heaviside}\left (t -1\right ) \sinh \left (t -1\right )^{2}}{2}
\]
✓ Solution by Mathematica
Time used: 0.046 (sec). Leaf size: 54
\[
x(t)\to \frac {1}{8} e^{-2 (t+1)} \left (\left (e^2-e^{2 t}\right )^2 \theta (1-t)+\left (e^2-1\right ) \left (e^{4 t}-e^2\right )\right )
\]
16.1.2 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(x(t),t$2)-4*x(t)=1-Heaviside(t-1),x(0) = 0, D(x)(0) = 0],x(t), singsol=all)
DSolve[{x''[t]-4*x[t]==1-UnitStep[t-1],{x[0]==0,x'[0]==0}},x[t],t,IncludeSingularSolutions -> True]