Internal problem ID [6503]
Internal file name [OUTPUT/5751_Sunday_June_05_2022_03_52_45_PM_9236993/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 7. Laplace Transforms. Section 7.5 The Unit Step and Impulse Functions. Page
303
Problem number: 7(b).
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`]]
\[ \boxed {L i^{\prime }+R i=E_{0} \delta \left (t \right )} \] With initial conditions \begin {align*} [i \left (0\right ) = 0] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} i^{\prime } + p(t)i &= q(t) \end {align*}
Where here \begin {align*} p(t) &=\frac {R}{L}\\ q(t) &=\frac {E_{0} \delta \left (t \right )}{L} \end {align*}
Hence the ode is \begin {align*} i^{\prime }+\frac {R i}{L} = \frac {E_{0} \delta \left (t \right )}{L} \end {align*}
The domain of \(p(t)=\frac {R}{L}\) is \[
\{-\infty
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (i\right ) =Y(s) \] Taking the Laplace transform of the ode
and using the relations that \begin {align*} \mathcal {L}\left (i^{\prime }\right )&= s Y(s) - i \left (0\right ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} L Y \left (s \right ) s -L i \left (0\right )+R Y \left (s \right ) = E_{0}\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} L Y \left (s \right ) s +R Y \left (s \right ) = E_{0} \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {E_{0}}{L s +R} \end {align*}
Taking inverse Laplace transform gives \[ \mathcal {L}^{-1}\left (\frac {E_{0}}{L s +R}\right ) = \frac {E_{0} {\mathrm e}^{-\frac {R t}{L}}}{L} \] The solution(s) found are the following \begin{align*}
\tag{1} i &= \frac {E_{0} {\mathrm e}^{-\frac {R t}{L}}}{L} \\
\end{align*} Verification of solutions
\[
i = \frac {E_{0} {\mathrm e}^{-\frac {R t}{L}}}{L}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [L i^{\prime }+R i=E_{0} \mathit {Dirac}\left (t \right ), i \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & i^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & i^{\prime }=\frac {-R i+E_{0} \mathit {Dirac}\left (t \right )}{L} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} i\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & i^{\prime }=-\frac {R i}{L}+\frac {E_{0} \mathit {Dirac}\left (t \right )}{L} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} i\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & i^{\prime }+\frac {R i}{L}=\frac {E_{0} \mathit {Dirac}\left (t \right )}{L} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (i^{\prime }+\frac {R i}{L}\right )=\frac {\mu \left (t \right ) E_{0} \mathit {Dirac}\left (t \right )}{L} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (i \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (i^{\prime }+\frac {R i}{L}\right )=i^{\prime } \mu \left (t \right )+i \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=\frac {\mu \left (t \right ) R}{L} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{\frac {R t}{L}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (i \mu \left (t \right )\right )\right )d t =\int \frac {\mu \left (t \right ) E_{0} \mathit {Dirac}\left (t \right )}{L}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & i \mu \left (t \right )=\int \frac {\mu \left (t \right ) E_{0} \mathit {Dirac}\left (t \right )}{L}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} i \\ {} & {} & i=\frac {\int \frac {\mu \left (t \right ) E_{0} \mathit {Dirac}\left (t \right )}{L}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{\frac {R t}{L}} \\ {} & {} & i=\frac {\int \frac {{\mathrm e}^{\frac {R t}{L}} E_{0} \mathit {Dirac}\left (t \right )}{L}d t +c_{1}}{{\mathrm e}^{\frac {R t}{L}}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & i=\frac {\frac {E_{0} \mathit {Heaviside}\left (t \right )}{L}+c_{1}}{{\mathrm e}^{\frac {R t}{L}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & i=\frac {{\mathrm e}^{-\frac {R t}{L}} \left (E_{0} \mathit {Heaviside}\left (t \right )+L c_{1} \right )}{L} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} i \left (0\right )=0 \\ {} & {} & 0=\frac {L c_{1} +E_{0}}{L} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {E_{0}}{L} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {E_{0}}{L}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & i=\frac {E_{0} {\mathrm e}^{-\frac {R t}{L}} \left (\mathit {Heaviside}\left (t \right )-1\right )}{L} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & i=\frac {E_{0} {\mathrm e}^{-\frac {R t}{L}} \left (\mathit {Heaviside}\left (t \right )-1\right )}{L} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 1.562 (sec). Leaf size: 17
\[
i \left (t \right ) = \frac {{\mathrm e}^{-\frac {R t}{L}} E_{0}}{L}
\]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 26
\[
i(t)\to \frac {\text {E0} (\theta (t)-\theta (0)) e^{-\frac {R t}{L}}}{L}
\]
24.5.2 Solving as laplace ode
24.5.3 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([L*diff(i(t),t)+R*i(t)=E__0*Dirac(t),i(0) = 0],i(t), singsol=all)
DSolve[{L*i'[t]+R*i[t]==E0*DiracDelta[t],{i[0]==0}},i[t],t,IncludeSingularSolutions -> True]