Laplace transform #1 - Q&A

Question 1 Solve the differential equation by variation of parameters yll + y = sin z Answer y"ty = sin x Complementary solution Auxiliary Equation is m²+1=0. m=i,-i Yc = C₁ cos x + C₂ sin x Particular solution y₁ = cosx , y₂= sin x , g(x)=sin(x) W (y₁, y₂) = | cosx sinx = cos²x + sin²x = 1 | -sinx cosx W₁ = | 0 sinx| = -sin²x= -(1-cos2x) = - 1/2 + cos2x/2 | sinx cosx W₂ = | cosx 0| = sinx cosx = 1/2 (2sinxcosx) = 1/2 sin 2x | -sinx sinx u₁ = ∫ W₁/W dx = ∫(-1/2+1/2cos2x) dx = -1/2∫dx+1/2∫cos2x dx u₁ = -1/2 x + 1/2(1/2 sin2x) = -1/2x + 1/4 sin 2x u₂ = ∫ W₂/W dx = ∫(1/2sin2x) dx = 1/2∫(sin2x)dx = 1/2(-1/2 cos2x) u₂ = -1/4 cos2x Yp = u₁y₁ + u₂y₂ Yp = (-1/2x + 1/4 sin2x)(cosx) + (-1/4 cos2x) (sin x) Yp = -1/2x cosx + 1/4 sin 2x cos x - 1/4 cos2x sinx Note: you can simplify above yp by applying appropriate trigonometric identities. However, Remember that this is NOT Required. It is optional. General solution y = yc + yp y = c₁ cosx + c₂ sin x - 1/2 x cos x + 1/4 sin 2x cos x - 1/4 cos 2x sin x Question 2 Solve the given differential equation by separation of variables. ds/dr = kS Answer 1 ds/dr = kS ds/S = kdr integrating on both the side, In (S) = kr + cc is integration constant. => S = C x e^(kr) C is integration constant. Answer 2 ds/dr = kS ds/S = kdr d²S/S = 2bkdr In |S| = kr + C Question 3 Apply unknown coefficients to the following differential equation to solve it. y" + 7y' + 6y = 30 Answer 1 Take a look at the differential equation. y" + 7y' + 6y = 30...(1) (D² + 7D + 6) y = 30 where D = d/dx f(m) = 0 m² +7m+6=0 m² + 6m+m+6=0 m(m+6) +1(m+6) = 0 (m+6) (m+1) = 0 m = -1,-6 yc = c₁e^(-x) + c₂e^(-6x) Answer 2 So, yp = Ax + B yp' = A, yp'' = 0 0 + 7A + 6(Ax+B)= 30 (7A+6B) + 6Ax = 30 6A = 0 7A+7B = 0 A = 0 and B = 5 (A = 0), (7(0) + 6B = 30), (6B = 30), (B = 5): yp = 0(x) + 5 = 5 The general solution: y = yc + yp yc = c₁e^(-x) + c₂e^(-6x)+5 yc = c₁e^(-x) + c₂e^(-6x) Question 4 Find the inverse Laplace Transform from the function below F(s) = 7/((s+11)(s+12)) Answer Step 1 We have to find inverse laplace transformation of given function. Step 2 Now, F(s) = 7/((s+11)(s+12)) Now, F(s) = 7*(1/((s+11)(s+12))) F(s) = 7*(1/(s+11) - 1/(s+12)) Therefore, L^(-1)F = L^(-1)[7/(s+11) - 7/(s+12)] = L^(-1) [7/(s+11)] - L^(-1) [7/(s+12)] = 7e^(-11t)- 7e^(-12t) Hence, L^(-1) [7/((s+11)(s+12))] = 7e^(-11t) - 7e^(-12t) Question 5 The Laplace transform of u(t - 2) is (a) 1/s - 2 (b) 1/s - 2/s (c) e^(-2s)d/s (d) e^(-2s)/s Answer 1 We know that: L(u(t)) = 1/s By t shifting theorem L(u(t-2)) = e^(-2s)L(u(t)) = e^(-2s)/s Answer would be (d) e^(-2s)/s Answer 2 To solve the problem, we'll start by recalling the definition of the unit step function. The unit step function, denoted as u(t), is defined as follows: u(t) = { 0 if t < 0 1 if t ≥ 0 Now, we are given the Laplace transform of u(t - 2) and we need to determine its value. We'll denote the Laplace transform operator as L. The Laplace transform of a function f(t) is given by: L{f(t)} = F(s) = ∫(0 to ∞)f(t)e^(-st)dt Applying the Laplace transform to u(t - 2), we have: L{u(t - 2)} = U(s) = ∫(0 to ∞)u(t-2)e^(-st)dt Since u(t-2) is a shifted version of the unit step function, it will be zero for t < 2 and one for t≥ 2. Therefore, we can rewrite the integral limits as follows: U(s) = ∫(2 to ∞)e^(-st)dt To evaluate this integral, we'll perform a change of variables. Let u = st, which implies du = sdt. Adjusting the integral limits accordingly, we get: U(s) = ∫(2s to ∞)e^(-u)du/s Now, we can solve this integral: U(s) = [-e^(-u)] (2s to ∞)/s Substituting the limits: U(s) = (-e^(-∞) + e^(-2s))/s Since e^(-∞) approaches zero, we can simplify further: U(s) = e^(-2s)/s Therefore, the Laplace transform of u(t - 2) is: L{u(t-2)} = e^(-2s)/s So, the correct option is (d) e^(-2s)/s Answer 3 The Laplace transform of u(t - 2) can be solved as follows: Using the definition of the unit step function, u(t) = 1 for t≥ 0 and u(t) = 0 for t < 0. Therefore, u(t-2) = 1 for t≥ 2 and u(t-2) = 0 for t < 2. The Laplace transform of u(t - 2) can be computed as: L{u(t-2)} = ∫(0 to ∞)u(t - 2)e^(-st)dt Since u(t-2) = 1 for t ≥ 2, we can rewrite the integral as: L{u(t-2)} = ∫(2 to ∞) e^(-st)dt Integrating the expression, we get: L{u(t-2)} = [-e^(-st)/s](2 to ∞) Substituting the limits of integration, we have: L{u(t-2)} = (-e^(-∞)/s) - (-e^(-2s)/s) Since e^(-s∞) = 0 (as e^(-∞) = 0), the expression simplifies to: L{u(t-2)} = - (-e^(-2s)/s) Therefore, the Laplace transform of u(t - 2) is e^(-2s)/s Hence, the correct option is (d) e^(-2s)/s