Internal problem ID [11529]
Internal file name [OUTPUT/10512_Thursday_May_18_2023_04_21_40_AM_82357209/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 3.4 Impulsive sources. Exercises page
173
Problem number: 10.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {x^{\prime \prime }+4 x=\frac {\left (t -5\right ) \operatorname {Heaviside}\left (t -5\right )}{5}+\left (2-\frac {t}{5}\right ) \operatorname {Heaviside}\left (-10+t \right )} \] 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 &=\frac {\left (10-t \right ) \operatorname {Heaviside}\left (-10+t \right )}{5}+\frac {\left (t -5\right ) \operatorname {Heaviside}\left (t -5\right )}{5} \end {align*}
Hence the ode is \begin {align*} x^{\prime \prime }+4 x = \frac {\left (10-t \right ) \operatorname {Heaviside}\left (-10+t \right )}{5}+\frac {\left (t -5\right ) \operatorname {Heaviside}\left (t -5\right )}{5} \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}^{-10 s}+{\mathrm e}^{-5 s}}{5 s^{2}}\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}^{-10 s}+{\mathrm e}^{-5 s}}{5 s^{2}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {{\mathrm e}^{-10 s}-{\mathrm e}^{-5 s}}{5 s^{2} \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}^{-10 s}-{\mathrm e}^{-5 s}}{5 s^{2} \left (s^{2}+4\right )}\right )\\ &= -\frac {\operatorname {Heaviside}\left (-10+t \right ) \left (-20+2 t -\sin \left (-20+2 t \right )\right )}{40}+\frac {\operatorname {Heaviside}\left (t -5\right ) \left (2 t -10-\sin \left (2 t -10\right )\right )}{40} \end {align*}
Converting the above solution to piecewise it becomes \[
x = \left \{\begin {array}{cc} 0 & t <5 \\ \frac {t}{20}-\frac {1}{4}-\frac {\sin \left (2 t -10\right )}{40} & t <10 \\ \frac {1}{4}+\frac {\sin \left (-20+2 t \right )}{40}-\frac {\sin \left (2 t -10\right )}{40} & 10\le t \end {array}\right .
\] Simplifying the solution gives \[
x = \frac {\left (\left \{\begin {array}{cc} 0 & t <5 \\ 2 t -10-\sin \left (2 t -10\right ) & t <10 \\ 10+\sin \left (-20+2 t \right )-\sin \left (2 t -10\right ) & 10\le t \end {array}\right .\right )}{40}
\]
The solution(s) found are the following \begin{align*}
\tag{1} x &= \frac {\left (\left \{\begin {array}{cc} 0 & t <5 \\ 2 t -10-\sin \left (2 t -10\right ) & t <10 \\ 10+\sin \left (-20+2 t \right )-\sin \left (2 t -10\right ) & 10\le t \end {array}\right .\right )}{40} \\
\end{align*} Verification of solutions
\[
x = \frac {\left (\left \{\begin {array}{cc} 0 & t <5 \\ 2 t -10-\sin \left (2 t -10\right ) & t <10 \\ 10+\sin \left (-20+2 t \right )-\sin \left (2 t -10\right ) & 10\le t \end {array}\right .\right )}{40}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }+4 x=\frac {\left (10-t \right ) \mathit {Heaviside}\left (-10+t \right )}{5}+\frac {\left (t -5\right ) \mathit {Heaviside}\left (t -5\right )}{5}, 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & x^{\prime \prime }=-4 x+\frac {\mathit {Heaviside}\left (t -5\right ) t}{5}-\mathit {Heaviside}\left (t -5\right )+2 \mathit {Heaviside}\left (-10+t \right )-\frac {\mathit {Heaviside}\left (-10+t \right ) t}{5} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} x\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & x^{\prime \prime }+4 x=\frac {\mathit {Heaviside}\left (t -5\right ) t}{5}-\mathit {Heaviside}\left (t -5\right )+2 \mathit {Heaviside}\left (-10+t \right )-\frac {\mathit {Heaviside}\left (-10+t \right ) t}{5} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2 \,\mathrm {I}, 2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )=\cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=\sin \left (2 t \right ) \\ \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} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+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 )=\frac {\mathit {Heaviside}\left (t -5\right ) t}{5}-\mathit {Heaviside}\left (t -5\right )+2 \mathit {Heaviside}\left (-10+t \right )-\frac {\mathit {Heaviside}\left (-10+t \right ) t}{5}\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} \cos \left (2 t \right ) & \sin \left (2 t \right ) \\ -2 \sin \left (2 t \right ) & 2 \cos \left (2 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=-\frac {\cos \left (2 t \right ) \left (\int \sin \left (2 t \right ) \left (\left (10-t \right ) \mathit {Heaviside}\left (-10+t \right )+\left (t -5\right ) \mathit {Heaviside}\left (t -5\right )\right )d t \right )}{10}+\frac {\sin \left (2 t \right ) \left (\int \cos \left (2 t \right ) \left (\left (10-t \right ) \mathit {Heaviside}\left (-10+t \right )+\left (t -5\right ) \mathit {Heaviside}\left (t -5\right )\right )d t \right )}{10} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\frac {\left (\cos \left (20\right ) \sin \left (2 t \right )-\sin \left (20\right ) \cos \left (2 t \right )-2 t +20\right ) \mathit {Heaviside}\left (-10+t \right )}{40}+\frac {\mathit {Heaviside}\left (t -5\right ) \left (-\frac {\sin \left (2 t \right ) \cos \left (10\right )}{2}+\frac {\cos \left (2 t \right ) \sin \left (10\right )}{2}+t -5\right )}{20} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+\frac {\left (\cos \left (20\right ) \sin \left (2 t \right )-\sin \left (20\right ) \cos \left (2 t \right )-2 t +20\right ) \mathit {Heaviside}\left (-10+t \right )}{40}+\frac {\mathit {Heaviside}\left (t -5\right ) \left (-\frac {\sin \left (2 t \right ) \cos \left (10\right )}{2}+\frac {\cos \left (2 t \right ) \sin \left (10\right )}{2}+t -5\right )}{20} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} \cos \left (2 t \right )+c_{2} \sin \left (2 t \right )+\frac {\left (\cos \left (20\right ) \sin \left (2 t \right )-\sin \left (20\right ) \cos \left (2 t \right )-2 t +20\right ) \mathit {Heaviside}\left (-10+t \right )}{40}+\frac {\mathit {Heaviside}\left (t -5\right ) \left (-\frac {\sin \left (2 t \right ) \cos \left (10\right )}{2}+\frac {\cos \left (2 t \right ) \sin \left (10\right )}{2}+t -5\right )}{20} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=-2 c_{1} \sin \left (2 t \right )+2 c_{2} \cos \left (2 t \right )+\frac {\left (2 \cos \left (20\right ) \cos \left (2 t \right )+2 \sin \left (20\right ) \sin \left (2 t \right )-2\right ) \mathit {Heaviside}\left (-10+t \right )}{40}+\frac {\left (\cos \left (20\right ) \sin \left (2 t \right )-\sin \left (20\right ) \cos \left (2 t \right )-2 t +20\right ) \mathit {Dirac}\left (-10+t \right )}{40}+\frac {\mathit {Dirac}\left (t -5\right ) \left (-\frac {\sin \left (2 t \right ) \cos \left (10\right )}{2}+\frac {\cos \left (2 t \right ) \sin \left (10\right )}{2}+t -5\right )}{20}+\frac {\mathit {Heaviside}\left (t -5\right ) \left (-\cos \left (2 t \right ) \cos \left (10\right )-\sin \left (2 t \right ) \sin \left (10\right )+1\right )}{20} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=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} \\ {} & {} & x=\frac {\left (\cos \left (20\right ) \sin \left (2 t \right )-\sin \left (20\right ) \cos \left (2 t \right )-2 t +20\right ) \mathit {Heaviside}\left (-10+t \right )}{40}+\frac {\mathit {Heaviside}\left (t -5\right ) \left (-\frac {\sin \left (2 t \right ) \cos \left (10\right )}{2}+\frac {\cos \left (2 t \right ) \sin \left (10\right )}{2}+t -5\right )}{20} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {\left (\cos \left (20\right ) \sin \left (2 t \right )-\sin \left (20\right ) \cos \left (2 t \right )-2 t +20\right ) \mathit {Heaviside}\left (-10+t \right )}{40}+\frac {\mathit {Heaviside}\left (t -5\right ) \left (-\frac {\sin \left (2 t \right ) \cos \left (10\right )}{2}+\frac {\cos \left (2 t \right ) \sin \left (10\right )}{2}+t -5\right )}{20} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.594 (sec). Leaf size: 43
\[
x \left (t \right ) = \frac {\operatorname {Heaviside}\left (t -10\right ) \sin \left (2 t -20\right )}{40}-\frac {\operatorname {Heaviside}\left (t -5\right ) \sin \left (2 t -10\right )}{40}+\frac {\left (-2 t +20\right ) \operatorname {Heaviside}\left (t -10\right )}{40}+\frac {\left (t -5\right ) \operatorname {Heaviside}\left (t -5\right )}{20}
\]
✓ Solution by Mathematica
Time used: 0.085 (sec). Leaf size: 55
\[
x(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{40} (2 (t-5)+\sin (10-2 t)) & 517.7.2 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)=1/5*(t-5)*Heaviside(t-5)+(1-1/5*(t-5))*Heaviside(t-10),x(0) = 0, D(x)(0) = 0],x(t), singsol=all)
DSolve[{x''[t]+4*x[t]==1/5*(t-5)*UnitStep[t-5]+(1-1/5*(t-5))*UnitStep[t-10],{x[0]==0,x'[0]==0}},x[t],t,IncludeSingularSolutions -> True]