Find the Wronskian?

Question

Find the Wronskian of e^t*sin(t), e^t*cos(t).

Answer: -e^2t

fg'-gf'

(e^t*sin(t))(-e^t*sin(t)*cos(t)+e^t*cos(t))-(e^t*cos(t))(e^t*cos(t)+e^t*sin(t))

How do I simplify this?

Answer 1

A quick way: [f(x) e^x]' = e^x [f(x) + f'(x)].

 e^t*sin(t)[e^t*cos(t)]' - [e^t*sin(t)]' e^t*cos(t)

= e^t*sin(t) e^t (cos(t)-sin(t)) - e^t (sin(t) + cos(t)) e^t*cos(t)

= e^2t (sin(t)cos(t)-sin^2(t) - sin(t)cos(t)-cos^2(t))

= -e^2t

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 Attn:

Your work contains small error.

(e^t*sin(t))(-e^t*sin(t)*cos(t)+e^t*cos(t))-(e^t*cos(t))(e^t*cos(t)+e^t*sin(t))

= -e^2t [sin^2(t) - sin(t)cos(t) + cos^2(t) + cos(t)sin(t)]

= -e^2t