Find the derivative of the integral (limits 1 to 3x) ds [(s^(2)+7)/(s^2+4)] + in

Question

Explanation / Answer

Apply L to both sides: [s^2 Y(s) - 0s - (-7)] + 4 * -(d/ds) [s Y(s) - 0] - 4 Y(s) = 0 ==> (s^2 - 4) Y(s) + 7 - 4 [Y(s) + s Y'(s)] = 0 ==> (s^2 - 8) Y(s) + 7 - 4s Y'(s) = 0 ==> Y'(s) + (-s/4 + 2/s) Y(s) = (7/4)/s. Multiply both sides by the integrating factor e^[? (-s/4 + 2/s) ds] = s^2 e^(-s^2/8): s^2 e^(-s^2/8) Y'(s) + e^(-s^2/8) (-s^3/4 + 2s) Y(s) = (7s/4) e^(-s^2/8) ==> (d/ds) [s^2 e^(-s^2/8) * Y(s)] = (7s/4) e^(-s^2/8). Integrate both sides: s^2 e^(-s^2/8) * Y(s) = -7e^(-s^2/8) + C ==> Y(s) = -7/s^2 + Ce^(s^2/8) / s^2. However, we know that the Laplace transform of a function always tends to 0 as s?8. Since lim(s?8) e^(s^2/8) / s^2 = 8, we must have C = 0. So, Y(s) = -7/s^2, and inverting yields y(t) = -7t.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.