24.6 problem 7(c)

Internal problem ID [6504]
Internal file name [OUTPUT/5752_Sunday_June_05_2022_03_52_48_PM_16266900/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(c).
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`]]

\[ \boxed {L i^{\prime }+R i=E_{0} \sin \left (\omega t \right )} \] With initial conditions \begin {align*} [i \left (0\right ) = 0] \end {align*}

24.6.1 Existence and uniqueness analysis

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} \sin \left (\omega t \right )}{L} \end {align*}

Hence the ode is \begin {align*} i^{\prime }+\frac {R i}{L} = \frac {E_{0} \sin \left (\omega t \right )}{L} \end {align*}

The domain of \(p(t)=\frac {R}{L}\) is \[ \{-\infty

24.6.2 Solving as laplace ode

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 ) = \frac {E_{0} \omega }{\omega ^{2}+s^{2}}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} L Y \left (s \right ) s +R Y \left (s \right ) = \frac {E_{0} \omega }{\omega ^{2}+s^{2}} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {E_{0} \omega }{\left (\omega ^{2}+s^{2}\right ) \left (L s +R \right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {E_{0} \omega L}{\left (\omega ^{2} L^{2}+R^{2}\right ) \left (s +\frac {R}{L}\right )}-\frac {\left (L \sqrt {-\omega ^{2}}-R \right ) E_{0} \omega }{2 \left (\omega ^{2} L^{2}+R^{2}\right ) \sqrt {-\omega ^{2}}\, \left (s -\sqrt {-\omega ^{2}}\right )}+\frac {\left (-L \sqrt {-\omega ^{2}}-R \right ) E_{0} \omega }{2 \left (\omega ^{2} L^{2}+R^{2}\right ) \sqrt {-\omega ^{2}}\, \left (s +\sqrt {-\omega ^{2}}\right )} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {E_{0} \omega L}{\left (\omega ^{2} L^{2}+R^{2}\right ) \left (s +\frac {R}{L}\right )}\right ) &= \frac {E_{0} \omega L \,{\mathrm e}^{-\frac {R t}{L}}}{\omega ^{2} L^{2}+R^{2}}\\ \mathcal {L}^{-1}\left (-\frac {\left (L \sqrt {-\omega ^{2}}-R \right ) E_{0} \omega }{2 \left (\omega ^{2} L^{2}+R^{2}\right ) \sqrt {-\omega ^{2}}\, \left (s -\sqrt {-\omega ^{2}}\right )}\right ) &= \frac {E_{0} {\mathrm e}^{i \operatorname {csgn}\left (i \omega \right ) \omega t} \left (-i R \,\operatorname {csgn}\left (i \omega \right )-L \omega \right )}{2 \omega ^{2} L^{2}+2 R^{2}}\\ \mathcal {L}^{-1}\left (\frac {\left (-L \sqrt {-\omega ^{2}}-R \right ) E_{0} \omega }{2 \left (\omega ^{2} L^{2}+R^{2}\right ) \sqrt {-\omega ^{2}}\, \left (s +\sqrt {-\omega ^{2}}\right )}\right ) &= \frac {E_{0} {\mathrm e}^{-i \operatorname {csgn}\left (i \omega \right ) \omega t} \left (i R \,\operatorname {csgn}\left (i \omega \right )-L \omega \right )}{2 \omega ^{2} L^{2}+2 R^{2}} \end {align*}

Adding the above results and simplifying gives \[ i=\frac {E_{0} \left (R \sin \left (\omega t \right )+L \omega \left (-\cos \left (\omega t \right )+{\mathrm e}^{-\frac {R t}{L}}\right )\right )}{\omega ^{2} L^{2}+R^{2}} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} i &= \frac {E_{0} \left (R \sin \left (\omega t \right )+L \omega \left (-\cos \left (\omega t \right )+{\mathrm e}^{-\frac {R t}{L}}\right )\right )}{\omega ^{2} L^{2}+R^{2}} \\ \end{align*}

Verification of solutions

\[ i = \frac {E_{0} \left (R \sin \left (\omega t \right )+L \omega \left (-\cos \left (\omega t \right )+{\mathrm e}^{-\frac {R t}{L}}\right )\right )}{\omega ^{2} L^{2}+R^{2}} \] Verified OK.

24.6.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [L i^{\prime }+R i=E_{0} \sin \left (\omega 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} \sin \left (\omega 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} \sin \left (\omega 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} \sin \left (\omega 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} \sin \left (\omega 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} \sin \left (\omega 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} \sin \left (\omega t \right )}{L}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} i \\ {} & {} & i=\frac {\int \frac {\mu \left (t \right ) E_{0} \sin \left (\omega 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} \sin \left (\omega 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} \left (-\frac {\omega \,{\mathrm e}^{\frac {R t}{L}} \cos \left (\omega t \right )}{\frac {R^{2}}{L^{2}}+\omega ^{2}}+\frac {R \,{\mathrm e}^{\frac {R t}{L}} \sin \left (\omega t \right )}{L \left (\frac {R^{2}}{L^{2}}+\omega ^{2}\right )}\right )}{L}+c_{1}}{{\mathrm e}^{\frac {R t}{L}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & i=\frac {c_{1} \left (\omega ^{2} L^{2}+R^{2}\right ) {\mathrm e}^{-\frac {R t}{L}}-E_{0} \left (L \omega \cos \left (\omega t \right )-R \sin \left (\omega t \right )\right )}{\omega ^{2} L^{2}+R^{2}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} i \left (0\right )=0 \\ {} & {} & 0=\frac {c_{1} \left (\omega ^{2} L^{2}+R^{2}\right )-E_{0} \omega L}{\omega ^{2} L^{2}+R^{2}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {E_{0} \omega L}{\omega ^{2} L^{2}+R^{2}} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {E_{0} \omega L}{\omega ^{2} L^{2}+R^{2}}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & i=\frac {E_{0} \left (\omega L \,{\mathrm e}^{-\frac {R t}{L}}-L \omega \cos \left (\omega t \right )+R \sin \left (\omega t \right )\right )}{\omega ^{2} L^{2}+R^{2}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & i=\frac {E_{0} \left (\omega L \,{\mathrm e}^{-\frac {R t}{L}}-L \omega \cos \left (\omega t \right )+R \sin \left (\omega t \right )\right )}{\omega ^{2} L^{2}+R^{2}} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

✓ Solution by Maple

Time used: 1.75 (sec). Leaf size: 44

dsolve([L*diff(i(t),t)+R*i(t)=E__0*sin(omega*t),i(0) = 0],i(t), singsol=all)
 

\[ i \left (t \right ) = \frac {\left (\omega L \,{\mathrm e}^{-\frac {R t}{L}}-L \cos \left (\omega t \right ) \omega +\sin \left (\omega t \right ) R \right ) E_{0}}{\omega ^{2} L^{2}+R^{2}} \]

✓ Solution by Mathematica

Time used: 0.114 (sec). Leaf size: 47

DSolve[{L*i'[t]+R*i[t]==E0*Sin[\[Omega]*t],{i[0]==0}},i[t],t,IncludeSingularSolutions -> True]
 

\[ i(t)\to \frac {\text {E0} \left (L \omega e^{-\frac {R t}{L}}-L \omega \cos (t \omega )+R \sin (t \omega )\right )}{L^2 \omega ^2+R^2} \]