3. Let's explore the role of the resistance R in an LCR circuit by setting R 0 in (1.6.1) and then comparing the result with the solution found in Fig. 1.21. The particular solution (1.6.2) remains unchanged. Determine the homogeneous solution and satisfy the initial conditions (1.6.3). Plot the result and compare it with Fig. 1.21. Answer: The results show a persistent oscillation around the mean value QCV. There is no decay in the oscillation as shown in Fig. 1.21.

If we use a linear approximation of e at a = 0 to approximate e-1/3, then we obtain: А. 4/3. В. 2/3 С. -1. D. 0.5. Е. 3/4. F. None of these answer.

Assignment 5: Problem 10 Match the equations of the surface with the graphs below and do all the subparts correctly with less explanations as they are connected with the same question.

8 4. Suppose these two pulses overlap in a medium. Note that this is a snapshot graph with x on the horizontal axis. 4 a. Fill in the table with the total displacement yr at x=0,1,...,8. (I've filled in a few.) 3 b. In the next column, write a C if there is destructive, and N if neither. constructive interference going on, D if graph below. c. Use the numbers from the table to draw the superposition of the two pulses in the blank 7 6 5 4 2 3 X 0 2 0 (a) Ут 0 (b) C/D Interference? N N N (c) 3 2 1 0 -1 ک حد تک 4 -2 2 1 0 F -1 -2 -3 -4 x= 0 1 2 3 4 5 6 7 8 NI -4 x= 0 1 2 3 4 5 6 7 8

1 (a). In the given waveform shown. Find the ripple factor y, if V1 = 8 and V2 = 18. [1 v(Volts) v2 t(seconds) 1 (b) Draw the circuit diagram for Full Wave Center Tap rectifier with a capacitor filter. [1 M ws bu

6.29) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.

15:26 E * ll all 80% i E pdf.2021 J s Exam /Math I/ 14-3- dy Q1/ Find dx 1. if y u3 and u = 4x2 — 2х + 5. 2. y = (x² + 4)²(2x3 – 1)3 3. ху + х — 2у —1 %3D0 1 3 2 4. y = = + + x2 x3 5. y = esin2x+cos3x cos x 6. У 7. у %3D tan-13x2 8. у = In sin3x 9. y = x23* 10. у %3D sec 16х II

3.2.4

Question 6 of 10 Solve the DE below: yʻ – y = 1+t(e'); y(0) = 2 Select the correct response: y = 0.5 – e' + t²e* y = –1+ e' + 0.5t²e y = -0.5 + e² +t²e' y = 1 – e' + 0.5t²e*