laplace transformation of
f'(t)+f(t)=integration of 0 to t (sin(t-u)f(u)) du when f(0)=1 can you tell me the step by step?

Question

laplace transformation of
f'(t)+f(t)=integration of 0 to t (sin(t-u)f(u)) du when f(0)=1 can you tell me the step by step?

anonymous · Answer

attachment

jusaquikie · Answer

Watch this video it should help you with a better explanation than i can type here. https://www.khanacademy.org/math/differential-equations/laplace-transform/properties-of-laplace-transform/v/laplace-transform-of-the-unit-step-function

anonymous · Answer

actually i figured this out. sY-f(0)+Y=1/(s^2-1) * Y

which in turn becomes Y(s-1-1/(s^2-1))=1
so Y= 1/(s-1-1/(s^2-1) so factorizing and clearing the RHS. I get 1/(s*(s^2+s+1)

applying inverse laplace will get me Y= 1/s-1/(s^2+s+1)