Internal problem ID [11523]
Internal file name [OUTPUT/10506_Thursday_May_18_2023_04_21_26_AM_62940236/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: 2.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "linear", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_linear, `class A`]]
\[ \boxed {x^{\prime }+3 x=\delta \left (t -1\right )+\operatorname {Heaviside}\left (t -4\right )} \] With initial conditions \begin {align*} [x \left (0\right ) = 1] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime } + p(t)x &= q(t) \end {align*}
Where here \begin {align*} p(t) &=3\\ q(t) &=\delta \left (t -1\right )+\operatorname {Heaviside}\left (t -4\right ) \end {align*}
Hence the ode is \begin {align*} x^{\prime }+3 x = \delta \left (t -1\right )+\operatorname {Heaviside}\left (t -4\right ) \end {align*}
The domain of \(p(t)=3\) is \[
\{-\infty
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (x\right ) =Y(s) \] 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 ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-x \left (0\right )+3 Y \left (s \right ) = {\mathrm e}^{-s}+\frac {{\mathrm e}^{-4 s}}{s}\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} s Y \left (s \right )-1+3 Y \left (s \right ) = {\mathrm e}^{-s}+\frac {{\mathrm e}^{-4 s}}{s} \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {{\mathrm e}^{-s} s +{\mathrm e}^{-4 s}+s}{s \left (s +3\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} s +{\mathrm e}^{-4 s}+s}{s \left (s +3\right )}\right )\\ &= \operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +3}+\frac {\operatorname {Heaviside}\left (t -4\right ) \left (1-{\mathrm e}^{-3 t +12}\right )}{3}+{\mathrm e}^{-3 t} \end {align*}
Converting the above solution to piecewise it becomes \[
x = \left \{\begin {array}{cc} {\mathrm e}^{-3 t} & t <1 \\ {\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3} & t <4 \\ {\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3}+\frac {1}{3}-\frac {{\mathrm e}^{-3 t +12}}{3} & 4\le t \end {array}\right .
\]
The solution(s) found are the following \begin{align*}
\tag{1} x &= \left \{\begin {array}{cc} {\mathrm e}^{-3 t} & t <1 \\ {\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3} & t <4 \\ {\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3}+\frac {1}{3}-\frac {{\mathrm e}^{-3 t +12}}{3} & 4\le t \end {array}\right . \\
\end{align*} Verification of solutions
\[
x = \left \{\begin {array}{cc} {\mathrm e}^{-3 t} & t <1 \\ {\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3} & t <4 \\ {\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3}+\frac {1}{3}-\frac {{\mathrm e}^{-3 t +12}}{3} & 4\le t \end {array}\right .
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }+3 x=\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right ), x \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=-3 x+\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right ) \\ \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}\hspace {3pt} \\ {} & {} & x^{\prime }+3 x=\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right ) \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (x^{\prime }+3 x\right )=\mu \left (t \right ) \left (\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right )\right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (x \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (x^{\prime }+3 x\right )=x^{\prime } \mu \left (t \right )+x \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=3 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{3 t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (x \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) \left (\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right )\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & x \mu \left (t \right )=\int \mu \left (t \right ) \left (\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right )\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {\int \mu \left (t \right ) \left (\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right )\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{3 t} \\ {} & {} & x=\frac {\int {\mathrm e}^{3 t} \left (\mathit {Dirac}\left (t -1\right )+\mathit {Heaviside}\left (t -4\right )\right )d t +c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{3} \mathit {Heaviside}\left (t -1\right )+\frac {{\mathrm e}^{3 t} \mathit {Heaviside}\left (t -4\right )}{3}-\frac {\mathit {Heaviside}\left (t -4\right ) {\mathrm e}^{12}}{3}+c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x=-\frac {{\mathrm e}^{-3 t +12} \mathit {Heaviside}\left (t -4\right )}{3}+\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +3}+\frac {\mathit {Heaviside}\left (t -4\right )}{3}+{\mathrm e}^{-3 t} c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=-\frac {{\mathrm e}^{-3 t +12} \mathit {Heaviside}\left (t -4\right )}{3}+\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +3}+\frac {\mathit {Heaviside}\left (t -4\right )}{3}+{\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=-\frac {{\mathrm e}^{-3 t +12} \mathit {Heaviside}\left (t -4\right )}{3}+\mathit {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +3}+\frac {\mathit {Heaviside}\left (t -4\right )}{3}+{\mathrm e}^{-3 t} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 5.422 (sec). Leaf size: 36
\[
x \left (t \right ) = \operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +3}-\frac {\operatorname {Heaviside}\left (t -4\right ) {\mathrm e}^{-3 t +12}}{3}+\frac {\operatorname {Heaviside}\left (t -4\right )}{3}+{\mathrm e}^{-3 t}
\]
✓ Solution by Mathematica
Time used: 0.201 (sec). Leaf size: 53
\[
x(t)\to \frac {1}{3} e^{-3 t} \left (3 e^3 \theta (t-1)+\left (e^{12}-e^{3 t}\right ) \theta (4-t)+e^{3 t}-e^{12}+3\right )
\]
17.1.2 Solving as laplace ode
17.1.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
dsolve([diff(x(t),t)+3*x(t)=Dirac(t-1)+Heaviside(t-4),x(0) = 1],x(t), singsol=all)
DSolve[{x'[t]+3*x[t]==DiracDelta[t-1]+UnitStep[t-4],{x[0]==1}},x[t],t,IncludeSingularSolutions -> True]