Final answer:
To solve the initial value problem y''+2y'+5y=0, y(0)=2, y'(0)=-1 using Laplace transform, we apply the transform to both sides of the equation and solve for Y(s). We then use partial fraction decomposition and inverse Laplace transform to obtain the solution.
Step-by-step explanation:
The given differential equation is a second-order linear homogeneous equation, which can be solved using Laplace transform. To solve the initial value problem y''+2y'+5y=0, y(0)=2, y'(0)=-1, we will apply the Laplace transform to both sides of the equation. Using the property L{y''(t)} = s^2Y(s) - sy(0) - y'(0) and L{y'(t)} = sY(s) - y(0), we have (s^2Y(s) - sy(0) - y'(0)) + 2(sY(s) - y(0)) + 5Y(s) = 0. Substituting the initial conditions y(0)=2 and y'(0)=-1, we get the equation (s^2+2s+5)Y(s) = 2s-5. Solving for Y(s), we have Y(s) = (2s-5)/(s^2+2s+5). To find y(t), we can use partial fraction decomposition and inverse Laplace transform to obtain the solution.