• Sep 16th 2006, 05:32 AM
action259
i'm having trouble with this step function question, i know it has to be done using laplace transforms but getting stuck any help would be appreciated

y"+2y = r(t) y(0)=y'(0)=0

r(t)= 1 0<t<1
0 otherwise

and this question i found the homogeneous solutions (e^x and e^2x) but am having trouble finding the particular solution, i'm been trying to use varation of parameters

x^2y"-2xy'+2y=1/x^2
• Sep 16th 2006, 06:12 AM
topsquark
Quote:

Originally Posted by action259
i'm having trouble with this step function question, i know it has to be done using laplace transforms but getting stuck any help would be appreciated

y"+2y = r(t) y(0)=y'(0)=0

r(t)= 1 0<t<1
0 otherwise

and this question i found the homogeneous solutions (e^x and e^2x) but am having trouble finding the particular solution, i'm been trying to use varation of parameters

x^2y"-2xy'+2y=1/x^2

Hint: Write r(t) in terms of the Heaviside step function. Specifically
r(t) = u(t) - u(t-1)
Where
u(t) = 0; t<0
u(t) = 1; t>0

-Dan