Laplace transform #10 - Q&A

Question 1 Use Laplace transform to evaluate the integral ∫(0 to ∞) e^(-2t) sin(2t)dt a) π/2 b) not defined c) 4s/(s²+4)² d) 4(s+2)/(s²+4s+8)² Answer Step 1 Here, the objective is to evaluate the given integral using Laplace transformation ∫(0 to ∞) e^(-2t) sin(2t)dt Step 2 According to the definition of Laplace transformation F(s) = L(f(t)) = ∫(0 to ∞) e^(-st) f(t)ds Compare the given integral with (1) f(t) = sin(2t) To evaluate the given integral, find the Laplace transformation of f(t) F(s) = L(f(t)) F(s) = L(sin 2t) Since L(sin(at)) = a/(s² + a²) Use L(tf(t)) = -d/ds (F(s)), Here f(t) = sin(2t) ⇒ F(s) = 2/(s²+4) Therefore F(s) = L(sin(2t)) = 2/(s²+4) F(s) = L(e^(-2t) sin(2t)) = 2/((s+2)²+4) F(s) = 2/(s²+4s+8) Step 3 Hence ∫(0 to ∞) e^(-2t) sin(2t)dt = 2/(s²+4s+8) Question 2 Find the inverse Laplace Tranformation by using convolution theorem for the function 1/(s³(s-5)) Answer Step 1 According to the given information, it is required to find the inverse of Laplace transform using convolution theorem. 1/(s³(s-5)) it is required to find L⁻¹{1/(s³(s-5))} Step 2 Solving further: let G(s) = 1/s³, H(s) = 1/(s-5) the inverse laplace of G(s) and H(s) are L⁻¹(G(s)) = L⁻¹{1/s³} = t²/2 = g(t) L⁻¹(H(s)) = L⁻¹{1/(s-5)} = e^(5t) = f(t) Step 3 The convolution theorem states that if G(s) and H(s) are the Laplace transform of g(t) and h(t) then, L{(gh)(t)} = G(s)H(s) ⇒ (gf)(t) = L⁻¹{G(s)H(s)} ⇒ g(t)*f(t) = L⁻¹{F(s)} = ∫(0 to t) f(t - u)g(u)du Step 4 Now solve the integral to find the inverse. ∫(0 to t) e^(5(t-u)) (u²/2)du = e^(5t) ∫(0 to t) e^(-5u) (u²/2)du = e^(5t)/2 ∫(0 to t) e^(-5u)u²du = e^(5t)/2 [-u²e^(-5u)/5 - 2ue^(-5u)/25 -2e^(-5u)/125] (from 0 to t) = e^(5t)/2 [-e^(-5t)t²/5 - 2e^(-5t) t/25 -2e^(-5t)/125 + 0 +0 + 2/125] = -t²/10 - t/25 - 1/125 + e^(5t)/125 ∫(0 to t) e^(5(t-u)) (u²/2)du = -t²/10 - t/25 - 1/125 + e^(5t)/125 Question 3 The laplace transform for ramp function with a time delay can be expressed as follows Answer Step 1 Let the time delay is T, Therefore, the ramp function will f(t) = (t-T)u(t - T) Where u(t-T) = { 1 t > T { 0 t < T Step 2 Let the Laplace transform of f(t) is F(s) The Laplace transform is therefore F(s) = ∫(0 to ∞) f(t)e^(-st)dt ⇒ F(s) = ∫(T to ∞) (t - T)e^(-st)dt let x=t-T ⇒ dx = dt Thus, we have F(s) = ∫(0 to ∞) xe^(-s(x+T)) dx F(s) = e^(-sT)∫(0 to ∞) xe^(-sx)dx F(s) = e^(-sT)/s² Question 4 Solve the initial value problem for r as a vector function of t. Differential equation: dr/dt = -ti - tj - tk Initial condition: r(0)=i+2j+3k Answer 1 r = ∫(-ti-tj-tk)dt = -t²i/2 - t²j/2 - t²k/2 + C; r(0) = 0i - 0j - 0k + C = i + 2j + 3k => C = i + 2j + 3k Result: r = (-t²/2 + 1)i + (-t²/2 + 2)j + (-t²/2 + 3)k Answer 2 Step 1: Integrate both sides of the differential equation with respect to t: ∫dr/dt dt = ∫(-ti-tj- tk)dt Step 2: Apply the integration rule for each component: r(t) = ∫-tidt + ∫-tjdt + ∫-tkdt Step 3: Evaluate the integrals: r(t) = -t²i/2 - t²j/2 - t²k/2 + c where c is the constant of integration. Step 4: Use the initial condition r(0) = i + 2j + 3k to determine the value of C: i + 2j + 3k = -(0)²i/2 - (0)²j/2 - (0)²k/2 + c Simplifying the equation: C = i +2j+3k Step 5: Substitute the value of C back into the solution: r(t) = -t²i/2 - t²j/2 - t²k/2 + i + 2j + 3k Simplifying further: r(t) = (1-t²/2)i + (2-t²/2)j + (3-t²/2)k Therefore, the solution to the initial value problem is: r(t) = (1-t²/2)i + (2-t²/2)j + (3-t²/2)k Answer 3 To solve the initial value problem for r as a vector function of t, with the given differential equation and initial condition, we can proceed as follows. The differential equation is given by: dr/dt = -ti-tj-tk where i, j, and k are the standard unit vectors in the Cartesian coordinate system. To solve this differential equation, we'll integrate both sides with respect to t. Integrating the right-hand side: ∫(-ti-tj-tk)dt = ∫-tidt- ∫tjdt - ∫tkdt ⇒ -t²i/2 - t²j/2 - t²k/2 + C where c is a constant vector of integration. Now, we need to find the constant vector c using the initial condition: r(0) = i + 2j + 3k Substituting t = 0 and r(0) into the solution expression, we get: i + 2j + 3k = -(0)²i/2 - (0)²j/2 - (0)²k/2 + C Simplifying, we find: i + 2j + 3k = C Therefore, the constant vector c is equal to i + 2j + 3k. Substituting this value back into the solution expression, we have: r(t) = -t²i/2 - t²j/2 - t²k/2 + (i + 2j + 3k) r(t) = -t²i/2 - t²j/2 - t²k/2 + (i + 2j + 3k) r(t) = (i-t²i/2) + (2j - t²j/2) + (3k-t²k/2) r(t) = i(1-t²/2) + j(2-t²/2) + k(3-t²/2) Hence, the solution to the initial value problem is: r(t) = (1-t²/2)i + (2-t²/2)j + (3-t²/2)k Answer 4 Step 1: First, let's separate the variables and integrate both sides of the differential equation: dr/dt = (-ti-j-k)dt ∫dr = ∫(-ti-j-k)dt r = ∫(-ti-j-k)dt Step 2: Next, we can integrate each component separately. Since the constant of integration wil be a vector, we can denote it as C: r = ∫(-ti)dt - ∫(j)dt - ∫(k)dt + C r = (-t²i/2) – (tj) – (tk) + C Step 3: Now, let's apply the initial condition r(0) = i + 2j + 3k to find the constant c: r(0) = (-(0)²i/2) – (0j) – (0k) + c i + 2j + 3k = C Therefore, the constant of integration C is i + 2j + 3k. Hence, the solution to the initial value problem is: r(t) = -t²i/2 - tj - tk + (i + 2j + 3k) We can simplify this further: r(t) = (-t²/2 + 1)i + (2 - t)j + (3 - t)k Question 5 Solve the given differential equation by separation of variables. dP/dt = P - P² Answer Solving the separable DE, first we multiply the whole equation by dt/(P-P²) to get dP/(P-P²) = dt ∫ dP/(P-P²) = ∫ dt Partial fractions can be used to solve the L.H.S integral, and we will format them as follows: 1/P(1-P) = A/P + B/(1-P) , where A and B are constants We can determine the values of A and B using the cover up rule First, cover up rule, we can get the values of A and B First, cover up P and substitute by P=0 to get A 1/(1-0) = A => A = 1 Then, cover up (1-P) and substitute by P=1 to get B 1/1 = B => B = 1 Reintroduce substitution into the equation to obtain ∫ (1/P + 1/(1-P))dP = ∫ dt => ln|P| - ln|1-P| = t + C1 ln|P/(1-P)| = t + C1 Exponentiating both sides, we get P/(1-P) = e^(t+C1) = e^(t)e^(C1) = Ce^(t) P = (1-P)Ce^(t) => P + PCe^(t) = Ce^(t) P(1+Ce^(t)) = Ce^(t) => P = Ce^(t)/(1+Ce^(t)) Result: P = Ce^(t)/(1+Ce^(t))