Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Show that when the voltage source in \(\left( {\bf{4}} \right)\) is of the form\({\bf{E(t) = }}{{\bf{E}}_{\bf{0}}}{\bf{sin}}\gamma {\bf{t}}\), then the steady-state solution \({{\bf{I}}_{\bf{p}}}\) is as given in the equation\(\left( {{\bf{10}}} \right)\).

Step-by-Step Solution

Answer

Find a particular solution of the equation\({\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{I}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{dI}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{I = }}\gamma {{\bf{E}}_{\bf{0}}}{\bf{cos}}\gamma {\bf{t}}\)by the method of undetermined coefficients. Assume that

 

\({{\bf{I}}_{\bf{p}}}{\bf{(t) = Acos}}\gamma {\bf{t + Bsin}}\gamma {\bf{t = }}\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} {\bf{sin(}}\gamma {\bf{t + \theta )}}\), where\({\bf{tan\theta   = A/B}}\).

Step 1:Using the method of undetermined coefficients

One needs to find a particular solution of\({\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{I}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{dI}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{I = }}\frac{{{\bf{dE}}}}{{{\bf{dt}}}}\)where\({\bf{E(t) = }}{{\bf{E}}_{\bf{0}}}{\bf{sin}}\gamma {\bf{t}}\).

So, our differential equation is\({\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{I}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{dI}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{I = }}\gamma {{\bf{E}}_{\bf{0}}}{\bf{cos}}\gamma {\bf{t}}\).

 

One will find a particular solution using the method of undetermined coefficients.

 

Assume that a particular solution has a form of\({{\bf{I}}_{\bf{p}}}{\bf{(t) = Acos}}\gamma {\bf{t + Bsin}}\gamma {\bf{t}}\).

 

Then one has\({\bf{I}}_{\bf{p}}^{}{\bf{'(t) =  - }}\gamma {\bf{Asin}}\gamma {\bf{t + }}\gamma {\bf{Bcos}}\gamma {\bf{t,}}\;{\bf{I}}_{\bf{p}}^{}{\bf{''(t) =  - }}{\gamma ^{\bf{2}}}{\bf{Acos}}\gamma {\bf{t - }}{\gamma ^{\bf{2}}}{\bf{Bsin}}\gamma {\bf{t}}\).

Step 2:Substituting the values in differential equation

Substituting this into the differential equation one will get:

 

\(\begin{aligned}{c}{\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{{\bf{I}}_{\bf{p}}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{d}}{{\bf{I}}_{\bf{p}}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{{\bf{I}}_{\bf{p}}}{\bf{ = L}}\left( {{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{Acos}}\gamma {\bf{t - }}{\gamma ^{\bf{2}}}{\bf{Bsin}}\gamma {\bf{t}}} \right){\bf{ + R( - }}\gamma {\bf{Asin}}\gamma {\bf{t + }}\gamma {\bf{Bcos}}\gamma {\bf{t)}}\\{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{(Acos}}\gamma {\bf{t + Bsin}}\gamma {\bf{t)}}\\{\bf{ = }}\left( {{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{LA + }}\gamma {\bf{RB + }}\frac{{\bf{A}}}{{\bf{C}}}} \right){\bf{cos}}\gamma {\bf{t + }}\left( {{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{LB - }}\,\gamma {\bf{RA + }}\frac{{\bf{B}}}{{\bf{C}}}} \right){\bf{sin}}\gamma {\bf{t}}\\{\bf{ = }}{{\bf{E}}_{\bf{0}}}\gamma {\bf{cos}}\gamma {\bf{t}}\end{aligned}\)

 

So,one has to solve a system;

 

\(\begin{aligned}{c}{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{LB - }}\gamma {\bf{RA + }}\frac{{\bf{B}}}{{\bf{C}}}{\bf{ = 0}}\\{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{LA + }}\gamma {\bf{RB + }}\frac{{\bf{A}}}{{\bf{C}}}{\bf{ = }}{{\bf{E}}_{\bf{0}}}\gamma \end{aligned}\)

Step 3: Substituting the values in the second equation

From the first equation, one has that\({\bf{A = }}\frac{{{\bf{B}}\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}}{{\gamma {\bf{RC}}}}...{\bf{(1)}}\),

Substituting this into the second equation one will get:

 

\(\begin{aligned}{c}\frac{{{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{LBC + }}{\gamma ^{\bf{4}}}{{\bf{L}}^{\bf{2}}}{\bf{B}}{{\bf{C}}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}{\bf{B + B - }}{\gamma ^{\bf{2}}}{\bf{LBC}}}}{{\gamma {\bf{R}}{{\bf{C}}^{\bf{2}}}}}{\bf{ = }}{{\bf{E}}_{\bf{0}}}\gamma \\\frac{{{\bf{B}}\left( {{\bf{ - 2}}{\gamma ^{\bf{2}}}{\bf{LC + }}{\gamma ^{\bf{4}}}{{\bf{L}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}{\bf{ + 1}}} \right)}}{{\gamma {\bf{R}}{{\bf{C}}^{\bf{2}}}}}{\bf{ = }}{{\bf{E}}_{\bf{0}}}\gamma \\\frac{{{\bf{B}}\left( {{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}} \right)}}{{\gamma {\bf{R}}{{\bf{C}}^{\bf{2}}}}}{\bf{ = }}{{\bf{E}}_{\bf{0}}}\gamma \\{\bf{B = }}\frac{{{{\bf{E}}_{\bf{0}}}{\gamma ^{\bf{2}}}{\bf{R}}{{\bf{C}}^{\bf{2}}}}}{{\left. {{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}} \right)}}\end{aligned}\)

Step 4:Substitute the values into the equation

Substituting the value for \({\bf{B}}\) into \(\left( {\bf{1}} \right)\)one will get;

 

\({\bf{A = }}\frac{{{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}}}{{\gamma {\bf{RC}}}}\frac{{{{\bf{E}}_{\bf{0}}}{\gamma ^{\bf{2}}}{\bf{R}}{{\bf{C}}^{\bf{2}}}}}{{\left. {{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}} \right)}} \Rightarrow {\bf{A = }}\frac{{{{\bf{E}}_{\bf{0}}}\gamma {\bf{C}}\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}}{{\left. {{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}} \right)}}\)

 

For a moment we won't substitute the values for \({\bf{A}}\)and\({\bf{B}}\). One will rewrite the particular solution for \({\bf{I}}\) as a sine function:

 

\(\begin{aligned}{c}{{\bf{I}}_{\bf{p}}}{\bf{(t) = Acos}}\gamma {\bf{t + Bsin}}\gamma {\bf{t}}\\{\bf{ = }}\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} \left( {\frac{{\bf{B}}}{{\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} }}{\bf{sin}}\gamma {\bf{t + }}\frac{{\bf{A}}}{{\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} }}{\bf{cos}}\gamma {\bf{t}}} \right)\\{\bf{ = }}\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} {\bf{sin(}}\gamma {\bf{t + \theta )}}\end{aligned}\)

 

Where\({\bf{tan\theta   = }}\frac{{\bf{A}}}{{\bf{B}}}\).

Step 5: Substitute the values for \({\bf{A,B}}\)

Now one can substitute values for \({\bf{A}}\)and\({\bf{B}}\). One has that;

 

\({\bf{tan\theta  = }}\frac{{\bf{A}}}{{\bf{B}}}{\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}\gamma {\bf{C}}\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}}{{{{\bf{E}}_{\bf{0}}}{\gamma ^{\bf{2}}}{\bf{R}}{{\bf{C}}^{\bf{2}}}}} \Leftrightarrow {\bf{tan\theta  = }}\frac{{{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}}}{{\gamma {\bf{RC}}}} \Leftrightarrow {\bf{tan\theta  = }}\frac{{{\bf{1/C - }}{\gamma ^{\bf{2}}}{\bf{L}}}}{{\gamma {\bf{R}}}}\)

 

One now only needs to calculate\(\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} \).

\(\begin{aligned}{c}\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} {\bf{ = }}\sqrt {{{\left( {\frac{{{{\bf{E}}_{\bf{0}}}\gamma {\bf{C}}\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}}{{{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}}}} \right)}^{\bf{2}}}{\bf{ + }}{{\left( {\frac{{{{\bf{E}}_{\bf{0}}}{\gamma ^{\bf{2}}}{\bf{R}}{{\bf{C}}^{\bf{2}}}}}{{{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}}}} \right)}^{\bf{2}}}} \\{\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}\gamma {\bf{C}}}}{{{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}}}\sqrt {{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + (}}\gamma {\bf{RC}}{{\bf{)}}^{\bf{2}}}} \\{\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}\gamma {\bf{C}}}}{{\sqrt {{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}} }}\\{\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}}}{{\sqrt {\frac{{{{\left( {{\bf{1 - }}{\gamma ^{\bf{2}}}{\bf{LC}}} \right)}^{\bf{2}}}{\bf{ + }}{\gamma ^{\bf{2}}}{{\bf{R}}^{\bf{2}}}{{\bf{C}}^{\bf{2}}}}}{{{\gamma ^{\bf{2}}}{{\bf{C}}^{\bf{2}}}}}} }}\\{\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}}}{{\sqrt {{{\left( {\frac{{\bf{1}}}{{\gamma {\bf{C}}}}{\bf{ - }}\gamma {\bf{L}}} \right)}^{\bf{2}}}{\bf{ + }}{{\bf{R}}^{\bf{2}}}} }}\\\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} {\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}}}{{\sqrt {{{\bf{R}}^{\bf{2}}}{\bf{ + }}{{\left( {\gamma {\bf{L - }}\frac{{\bf{1}}}{{\gamma {\bf{C}}}}} \right)}^{\bf{2}}}} }}\end{aligned}\)

Step 6: Steady-state solution

One has that the steady-state solution \({{\bf{I}}_{\bf{p}}}\) is given by;

 

\(\begin{aligned}{c}{{\bf{I}}_{\bf{p}}}{\bf{(t) = }}\sqrt {{{\bf{A}}^{\bf{2}}}{\bf{ + }}{{\bf{B}}^{\bf{2}}}} {\bf{sin(}}\gamma {\bf{t + \theta )}}\\{\bf{ = }}\frac{{{{\bf{E}}_{\bf{0}}}}}{{\sqrt {{{\bf{R}}^{\bf{2}}}{\bf{ + }}{{\left( {\gamma {\bf{L - }}\frac{{\bf{1}}}{{\gamma {\bf{C}}}}} \right)}^{\bf{2}}}} }}{\bf{sin(}}\gamma {\bf{t + \theta )}}\end{aligned}\)

 

Where\({\bf{tan\theta  = }}\frac{{\frac{{\bf{1}}}{{\bf{C}}}{\bf{ - }}{\gamma ^{\bf{2}}}{\bf{L}}}}{{\gamma {\bf{R}}}}\)which we wanted to show.