t^2 y'' − 4ty' + 6y = t^4*e^t , t > 0. Use variation of
parameters...
t^2 y'' − 4ty' + 6y = t^4*e^t , t > 0. Use variation of
parameters to find a particular solution given that y1 = t^2 and y2
= t^3 are a fundamental set of solutions to the corresponding
homogeneous equation
($4.6 Variation of Parameters): Solve the equations (a)–(c)
using method of variation of parameters.
(a) y''-6y+9y=8xe^3x
(b) y''-2y'+2y=e^x (secx)
(c) y''-2y'+y= (e^x)/x