Skip to main content

Solve ODE by Laplace transform y''+2y'+2y=5sinx

Short answer

Using Laplace transform gives Y = 5/[(s^2+1)(s^2+2s+2)], which by partial fractions becomes (-2s+1)/(s^2+1) + (2s+3)/((s+1)^2+1). Inverting yields y(x)= sin x - 2 cos x + e^{-x}(2 cos x + sin x).

Using Laplace transform, solve the initial value problem: y'' + 2y' + 2y = 5 sin x, y(0) = y'(0) = 0.

y″ + 2y′ + 2y = 5 sin x, y(0) = y′(0) = 0. Here x is the independent variable (use t in the transform).

Step 1: Transform. With Y = L[y]:

(s² + 2s + 2) Y = 5/(s² + 1)
Y = 5 / [ (s² + 1)(s² + 2s + 2) ]

Step 2: Partial fractions.

Y = (As + B)/(s² + 1) + (Cs + D)/((s + 1)² + 1)

Solving (compare coefficients) gives A = −2, B = 1, C = 2, D = 3:

Y = (−2s + 1)/(s² + 1) + (2s + 3)/((s + 1)² + 1)

Check of the numerator: (−2s + 1)(s² + 2s + 2) + (2s + 3)(s² + 1) = 5 ✓.

Step 3: Invert. Write (2s + 3) = 2(s + 1) + 1:

y = −2 cos x + sin x + e^(−x) (2 cos x + sin x)

Answer: y(x) = sin x − 2 cos x + e^(−x)(2 cos x + sin x)

Check: y(0) = −2 + 2 = 0 ✓. y′ = cos x + 2 sin x − e^(−x)(2 cos x + sin x) + e^(−x)(cos x − 2 sin x), so y′(0) = 1 − 2 + 1 = 0 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →