Laplace transform to solve initial value problem. Use Laplace transforms to solv

Question

Laplace transform to solve initial value problem.

Use Laplace transforms to solve the initial value problem: d2y/dt2 + 6 dy/dt + 5y = cos (3t); where y(0) = 0, dy/dt (0) = 0 Step one: Take the Laplace transforms of the LHS and RHS. You find: Y(s) = (y(t)) = Answer is a Expression Step two: Using partial fractions you can rewrite Y(s) as: Y(s) = As + B/s2 + 3 2 + C/s + 1 + D/s + 5, where A = B = C = D = Step three: State the solution of the initial value problem by finding the inverse Laplace transform of Y(s): y(t) = -1(Y(s)) = Answer is a Expression Laplace Transform table

Explanation / Answer

Y(s) = s/[(s^2+9)*(s^2+6s+5)] = s/[(s^2+9)*(s+1)*(s+5)] = (As+B)/(s^2+9) + C/(s+1) + D/(s+5) we get A = -3/170 B = -2/17 C = 1/40 D = -1/136 Y(s) = -(3s+20)/[170(s^2+9)] + 1/[40(s+1)] - 1/[136(s+5) ] apply inverse laplace y(t) = (-3/170) * Cos(3t) - (2/51) * Sin(3t) + (1/40) * e^(-t) - (1/136)*e^(-5t)

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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