ﬁnd the general solution of the given differential equation 1. 2y''+3y'+y=t^2 +3sint ﬁnd the solution of...

ﬁnd the general solution of the given differential equation

1. 2y''+3y'+y=t^2 +3sint

ﬁnd the solution of the given initial value problem

1. y''−2y'−3y=3te^2t, y(0) =1, y'(0) =0

2. y''−2y'+y=te^t +4, y(0) =1, y'(0) =1

Expert Solution

Colby Messinger answered 1 year ago

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