Final answer:
To solve the Cauchy-Euler equation t²y''(t)+4ty'(t)+2y(t)=0, we assume a solution of the form y(t) = t^r and find the characteristic equation. Solving the quadratic equation gives us r = -1 and r = -2. The general solution is y(t) = c1t^(-1) + c2t^(-2), and applying the initial conditions gives us c1 = -10 and c2 = 6.
Step-by-step explanation:
To solve the Cauchy-Euler equation t²y''(t)+4ty'(t)+2y(t)=0, we can assume a solution of the form y(t) = t^r. Differentiating twice yields y''(t) = r(r-1)t^(r-2). Substituting these expressions back into the original equation and dividing by t^r, we get the characteristic equation r(r-1) + 4r + 2 = 0. Solving this quadratic equation gives us r = -1 and r = -2. So the general solution is y(t) = c1t^(-1) + c2t^(-2), where c1 and c2 are constants.
Applying the initial conditions y(1) = -4 and y'(1) = 10, we can substitute t = 1 into the equation y(t) = c1t^(-1) + c2t^(-2) and its derivative y'(t) = -c1t^(-2) - 2c2t^(-3). This gives us -4 = c1 + c2 and 10 = -c1 - 2c2. Solving these equations simultaneously, we find c1 = -10 and c2 = 6. Therefore, the solution to the initial value problem is y(t) = -10/t + 6/t^2.