Find the general solution of , where y is a function of t.
Could someone help me out with this? I'm not sure how to go about solving for the general solution. All I've got is $\displaystyle dy/dt = 1-2y$.
Thanks!
alright so this is what I ended up trying, but I don't think it's correct. any pointers on where I went wrong?
$\displaystyle e^{2t}y'+2e^{2t}y = e^{2t}$
$\displaystyle (e^{2t}y)' = e^{2t}$
$\displaystyle e^{2t}y = \int e^{2t} dt$
$\displaystyle e^{2t}y = (e^{2t})/2 +c$
$\displaystyle y= 1/2 +c$
y' +2y =1
first find the characteristic roots
y'+2y=0
lambda=l
l^2+2=0
l=-2
l=+-sqrt(2)i
y=e^sqrt(2)i+e^-sqrt(2)i
expand with eulers formula get sin and cosine = yg, then
find particular solution
y=A
y'=0
plug into the question
2A=1
A=1/2
Now yp=1/2
y=yg+yp
Cheers