The characteristic equation of the differential equation is: r^(2)-10r+25=0 Factorize the characteristic equation: (r-5)^(2)=0 Determine the roots of the characteristic equation of the differential equation: r=5 If the characteristic equation has two real roots r_(1) and r_(2), then the general solution is y(t)=c_(1)e^(r_(1)t)+c_(2)e^(r_(2)t). If the characteristic equation has one real root r, then the general solution is y(t)=c_(1)e^(rt)+c_(2)te^(rt). Since this characteristic equation has one real root, the general solution is: y(t)=c_(1)e^(5t)+c_(2)te^(5t) A basis for the solution space is then made up by the coefficients of the variables c_(1) and c_(2) in the general solution: (e^(5t), te^(5t)) The solution space is then the span of the basis: Span (e^(5t), te^(5t)) Result: Basis: (e^(5t), te^(5t)) Solution space: Span (e^(5t), te^(5t))