RLC Circuit Homework - Solving 3rd Part with 2nd Order DE

Click For Summary
SUMMARY

The discussion focuses on solving the third part of an RLC circuit homework problem using a second-order differential equation, specifically y'' + ay' + b = 0. The user successfully solved the first two parts but found the third part, involving V = V_o sin wt, challenging. They propose using a particular integral of the form y = A sin wt + B cos wt and suggest leveraging the relationship between sine functions and complex exponentials, specifically V_0 sin wt = Im(V_0 exp(iwt)), to simplify the solution process.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with RLC circuit analysis
  • Knowledge of complex numbers and their applications in electrical engineering
  • Ability to differentiate and manipulate trigonometric functions
NEXT STEPS
  • Study the method of solving second-order differential equations with constant coefficients
  • Learn about the application of complex exponentials in circuit analysis
  • Explore techniques for finding particular integrals in differential equations
  • Research the relationship between real and imaginary components in differential equations
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in solving differential equations related to RLC circuits will benefit from this discussion.

bigevil
Messages
77
Reaction score
0

Homework Statement



2w32nlt.jpg


The Attempt at a Solution



Ok, first part, no problem, second part (steady state), solved in another thread. Both are pretty tedious, but doable.

However, I am quite stumped by the third part of this problem (V = V_o sin wt) .

The second part of the problem requires me to use a 2nd order differential equation (y'' + ay' + b = 0), and the solutions of the auxiliary equation can be saved for the third part.

For the third part, I choose a particular integral y = A sin wt + B cos wt. The last part arises essentially because the GS of the third part is y = exponent + exponent + A sin wt + B cos wt, and the first two terms become 0 when t is large. Then I can reexpress A sin wt + B cos wt into the form required.

I have just tried to go about the third part the standard way (differentiating and substituting the particular integral to find A and B, and then substituting again to find the constants in the complementary function), but it is extremely tedious (one page working just to find and simplify A or B) and I hope there is a shorter solution.
 
Physics news on Phys.org
There is a well-known trick. V_0 sin wt is the imaginary part of V_0 exp(iwt).
i.e. V_0 sin wt = Im( V_0 exp(iwt) )

Also, since the coefficients of the differential equation are real, real and imaginary part of the left and the right hand side of the equation behaves independently. Hence, the solution you are looking for is the imaginary part of the solution for V_0 exp(iwt).

Try to solve for V_0 exp(iwt). It simplifies the problem considerably.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
8K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
1
Views
2K
Replies
1
Views
9K
  • · Replies 11 ·
Replies
11
Views
5K