ESE 2026Q59
Q: The solution of differential equation y″ + 4y′ + 3y = eᵗ; y(0) = 0, y′(0) = 2 is
(a)y(t) = eᵗ/8 − 7e⁻³ᵗ/8 + 3e⁻ᵗ/4
(b)y(t) = eᵗ/8 − 5e⁻³ᵗ/8 + 3e⁻ᵗ/4
(c)y(t) = eᵗ/8 + 7e⁻³ᵗ/8 − 3e⁻ᵗ/4
(d)y(t) = eᵗ/8 + 7e⁻³ᵗ/8 + 3e⁻ᵗ/4
Answer pending verification