Originally Posted by
zhupolongjoe Solve:
2 (dy/dt)=2y+3t
So I divide through by 2:
(dy/dt)=y+(3/2)t
I take the homogenous yh=e^t
Then I make it
(dy/dt)-y=(3/2)t
So I guess a*t
Plug that in to get:
a-at=(3/2)t
So that
a(1-t)=(3/2)t
and then
a=(3/2)t/(1-t)
And multiplying that my t and combing with the homogeous I get:
y(t)=e^t+(3/2)t^2/(1-t)
but I've never seen an answer like this since usually it is a more complicated function with es and cosines....
Is this correct or where did I make a mistake?
Thanks.