Solve #y+3y'+2y=(t^2+1)e^-t#?
1 Answer
# y = t^3/9e^(-t) + t/3e^(-t) + Ce^(-t) #
Explanation:
We have:
# y + 3y' + 2y = (t^2 + 1)e^(-t) #
Which we can write as:
# 3y' + 3y = (t^2 + 1)e^(-t) #
# :. y' + y = 1/3(t^2 + 1)e^(-t) #
We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;
# dy/dx + P(x)y=Q(x) #
As the equation is already in this form, then the integrating factor is given by;
# I = exp(int \ P(x) \ dx #
# \ \ = exp(int 1 \ dt) #
# \ \ = exp(t) #
# \ \ = e^t #
And if we multiply the DE by this Integrating Factor,
# e^t y' + e^ty = 1/3(t^2 + 1) #
# :. d/dt(e^ty) = 1/3(t^2 + 1) #
This is now separable, so by "separating the variables" we get:
# e^ty = int \ 1/3(t^2 + 1) \ dt#
This is directly integrable, so doing so gives us:
# e^ty = 1/3(t^3/3 + t) + C#
Leading to the GS:
# y = t^3/9e^(-t) + t/3e^(-t) + Ce^(-t) #
