University:
North Central College
Course:
MATH 152 | Calculus II
Academic year:

2024
Views:

192

Pages:

4
Author:

nesikhiyawl0

Question 1 Use Theorem 7.4.3 to find the Laplace transform F(s) of the given periodic function. F(s)=? Answer 1 Given that f(t) is f(t) is periodic function with period 2a :. F(s) = L{f(t)} = 1/(1-e^(-2as)) ∫(0 to 2a) e^(-st) f(t)dt = 1/(1-e^(-2as)) [∫(0 to a) e^(-st).1 dt + ∫(a to 2a) e^(-st)(-1) dt] = 1/(1-e^(-2as)) [ -e^(-st)/s ](0 to a) - [ e^(-st)/s ](a to 2a) = 1/(1-e^(-2as)) [ -e^(-as)/s + 1/s - (-e^(-2as)/s + e^(-as)/s)] = 1/(1-e^(-2as)) [ -e^(-as)/s + 1/s + e^(-2as)/s - e^(-as)/s] = 1/(1-e^(-2as)) [ 1 - 2e^(-as) + e^(-2as)]/s = 1/(1-e^(-2as)) (1 - e^(-as))²/s = (1 - e^(-as))² / s(1 - e^(-as))(1 + e^(-as)) = (1 - e^(-as)) / s(1 + e^(-as)) = (1 - e^(-as)) / s(1 + e^(-as)) = 1/s (1 - e^(-as))/(1 + e^(-as)) Therefore, L{f(t)}= 1/s (1 - e^(-as))/(1 + e^(-as)) Answer 2 Given that f(t) = { 1 if 0 0: dy/dt + 4∫(0 to t) y(t-w)e^(-4w)dw = 3, y(0) = 0 a) Use convolution and Laplace transforms to find the Laplace transform of the solution. Y(s) = L{y(t))}-? b) Obtain the solution y(t). y(t) -? Answer Step 1 Given, dy/dt + 4∫(0 to t) y(t-w)e^(-4w)dw = 3, y(0) = 0 a) Taking Laplace transform of the above equation L(y') + 4L{∫(0 to t) y(t-w)e^(-4w)dw} = L (3) sy(s) - y(0) + 4y(s) e^(-4t) = 3/s sy(s) + 4y(s)(1/(s+4)) = 3/s y(s) = 3/s / (s + 4/(s+4)) = 3/s / (s(s+4)+4)/(s+4) = 3/s * (s+4) / (s^2 + 4s + 4) y(s) = 3(s+4) / s(s+2)² y(s) = 3(s+4)/(s(s+2)²) Step 2 Now take, 3(s+4)/(s(s+2)²) resolve into partial fractions 3(s+4)/(s(s+2)²) = A/s + B/(s+2) + C/(s+2)² 3(s+4)/(s(s+2)²) = (A(s+2)² + Bs(s+2) + Cs) / s(s+2)² 3(s+4) = A(s+2)² + Bs(s+2) + Cs Put s = 0 12 = 4A A = 3 Put s = -2 6 = 0 + 0 - 2C C = -3 Consider coefficients of 's2' 0 = A + B A = -B or B = -A B = -3 Put the values in the equation 3(s+4) / (s(s+2)²) = 3/s - 3/(s+2) - 3/(s+2)² Question 5 Find dy/dx by implicit differentiation. xe^y = x - y Answer xe^y = x - y Differentiate throughout with respect to x d/dx(xe^y) = d/dx(x-y) d/dx(x)e^y + x d/dx(e^y) = 1 - dy/dx (1)e^y + x(e^y(dy/dx)) = 1 - dy/dx e^y + xe^y dy/dx = 1 - dy/dx Add dy/dx on both sides e^y + xe^y dy/dx + dy/dx = 1 Subtract e^y from both sides xe^y dy/dx + dy/dx = 1 - e^y Take dy/dx common from the two terms in LHS (xe^y + 1)dy/dx = 1 - e^y Divides both sides by xe^y + 1 dy/dx = (1 - e^y)/(xe^y + 1) Result: dy/dx = (1 - e^y)/(xe^y + 1)