Hey guys I need help on 4 problems. Mind helping? f(t) = 2t + 1 if 0 t 2. Using

Question

Hey guys I need help on 4 problems. Mind helping?

f(t) = 2t + 1 if 0 t 2. Using the convolution theorem, find the inverse Laplace transform of the given function: Solve the following differential equation. y' - 6xy-2 = -2xy. Set up the appropriate form of a particular solution yp, but do not determine the values of the coefficients. (D - 1 )3(D2 + 4)y = xex + x2 cos(2x). Solve the initial value problem x2y" - 3xy' + 3y = 2x3, y(1) = 1, y'(1) = -2, given that yc = c1x + c2x3. Solve the initial value problem in the form of a series solution. Find only the first four non-zero terms of the series.

Explanation / Answer

6) let H(s) = 1/(s+2) G(s) = 2/[s^2+4] now h(t) = L^-1{1/(s+2)} = e^(-2t) g(t) = sin(2t) now using convolution theorem L^-1{1/(s+2) * 2/[s^2+4]} = L^-1{H(s) G(s)} = int from 0 to t [ e^(-2(t - T)) sin(2T) dT] = e^(-2t) * int from 0 to t [ e^(2T) sin(2T) dT] = e^(-2t) * [e^(2T)/8 * (2sin(2T) - 2cos(2T)] put limits from 0 to t = e^(-2t) * [e^(2t)/8 * (2sin(2t) - 2cos(2t)) - 1/8 *(0 - 2)] = (sin(2t) - cos(2t))/4 + e^(-2t)/4 7) y' - 6xy^-2 = -2xy or y' = -2xy + 6x/y^2 or y' = 2x[-y + 3/y^2] or dy*y^2/[-y^3+3] = 2x dx let -y^3 +3 = z -3y^2 dy = dz we get -dz/z * 1/3 = 2x dx now integrate -(1/3) ln(z) = x^2 + C or (-1/3) ln(3-y^3) = x^2 + C or ln(3 - y^3 ) = -3x^2 + K or 3 - y^3 = e^(-3x^2 + K) or y^3 = 3 - e^(-3x^2 + K) or y = [3 - e^(-3x^2 + K)]^(1/3) 8) Yp = 1/[(D-1)^3 (D^2+4)] [xe^x + x^2 cos(2x)]

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