Final answer:
The particular solution to y′′+10y′+25y= (5.5 e⁻⁵ᵗ)/(t²+1)n is y_p = (5.5e^(-5t))/(26(t^2 + 1)).
Step-by-step explanation:
To find a particular solution to this second-order linear homogeneous ordinary differential equation, we can use the method of undetermined coefficients.
Assume a particular solution of the form y_p = (Ce^(-5t))/(t^2 + 1) and solve for C.
Since the equation on the right-hand side is of the form (Ae^(-bt))/(t^2 + 1), we can assume a particular solution of the form y_p = (Ce^(-5t))/(t^2 + 1).
Plugging this into the equation and solving for C, we find that C = 5.5/26.
Therefore, a particular solution to the given differential equation is y_p = (5.5e^(-5t))/(26(t^2 + 1)).