Question

solve y''+2y'+5y=e-t sint y=(0) and y'(0)=1 using Laplace transform​

Answer 1

Originally Answered: How do you solve y'' + 2 y' + 5 y = e^-t sint when y(0) =0 and y'(0) = 1?

This is a linear differential equation, so the usual method of complementary function + particular integral will work.

COMPLEMENTARY FUNCTION

Start with the original differential equation but with the RHS being zero. So:

y'' + 2 y' + 5 y = 0

Write out the auxiliary equation:

m^2 + 2m + 5 = 0

Solve this. m = -1 + 2i or m = -1 - 2i

Write out the complementary function:

y = e^-t(Acos2t + Bsin2t)

PARTICULAR INTEGRAL

As the RHS is exponential x trig, assume the particular integral is likewise.

So y = e^-t(λ cos t + μ sin t)

Differentiate (using the product rule) to give y'.

Differentiate again (using the product rule again) to give y''.

Substitute all three of y, y' and y'' into the original differential equation.

Simplify by dividing through by e^-t.

Looking just at the cos t terms, write out an equation in λ and μ.

Looking just at the sin t terms, write out another equation in λ and μ.

Solve these two equations simultaneously to give λ and μ.

Substitute into y = e^-t(λ cos t + μ sin t) to give the particular integral.

INITIAL CONDITIONS

Having arrived at the general solution, use the fact that y(0) = 0.

This means that A + λ = 0, so A = -λ

Also differentiate the general solution to give y'.

Use the fact that y'(0) = 1 to work out B.

Substitute these values of A and B into the general solution to give the answer to the question.

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Answer 2

Answer:

Y(t) = 1/3 e^(-t) (sin(t) + sin(2 t))

Step-by-step explanation:

Apply the Laplace transformation Lt [f(t)](s) = integral_0^∞ f(t) e^(-s t) dt to both sides:

Lt [( d^2 Y(t))/( dt^2) + 2 ( dY(t))/( dt) + 5 Y(t)](s) = ℒ_t[e^(-t) sin(t)](s)

Evaluate Laplace transforms:

(s^2 + 2 s + 5) (Lt [Y(t)](s)) - 1 = 1/((s + 1)^2 + 1)

Solve for Lt [Y(t)](s):

Lt [Y(t)](s) = (s^2 + 2 s + 3)/((s^2 + 2 s + 2) (s^2 + 2 s + 5))

Decompose Lt [Y(t)](s) via partial fractions:

Lt [Y(t)](s) = 1/(3 (s^2 + 2 s + 2)) + 2/(3 (s^2 + 2 s + 5))

Take the inverse Laplace transform of both sides:

Y(t) = 1/3 e^(-t) (sin(t) + sin(2 t))