Substitute $y=e^{rx}$ to obtain the characteristic equation $r^2+2r-5=0$.
The quadratic formula gives $r=(-2\pm\sqrt{24})/2=-1\pm\sqrt6$. These are distinct roots, so the two exponential solutions are independent. Their arbitrary linear combination is the expression in **A**.
More Differential Equations questions
GATE 2016 EC (EC2) Q36– The equation dx/dt=-3x+2, x(0)=1, is solved by forward Euler. The largest time step that…
GATE 2016 EC (EC3) Q15– Consider y=y+2x-x2, y(0)=1, for 0 x<, with exact solution y(x)=x2+ex. At x=0.1, the…
GATE 2016 EC (EC3) Q36– The initial-value problem y+12y+36y=0, y(0)=3, y(0)=-36 has solution
GATE 2026 EC Q11– Consider the differential equation w = Aw, with w(t=0) = bmatrix1\\1bmatrix. If w(t) = etu…
GATE 2025 EC Q32– The function y(t) satisfies t2y''(t)-2ty'(t)+2y(t)=0, where y'(t) and y''(t) denote the…
GATE 2024 EC Q11– The general form of the complementary function of a differential equation is given by…