Question

use the fundamental formula to find a formula for y(x) in terms of x given that y''(x)= (e^(-x))*(cos(x)). I i'm a little bit mess up by second derivative. Thanks

Answer 1

this looks like a big fat pain
have to integrate by parts for \(y'=\int e^{-x}\cos(x)dx\)

Answer 2

i think just use in fundamental formula step by step to show up would be fine. i'm little bit mess up with second derivative

Answer 3

i don't know what "fundamental formula" means, unless it is
\[\int e^{ax}\cos(bx)dx=\frac{e^{ax}(a\cos(bx)+b\sin(bx))}{a^2+b^2}\]

Answer 4

if that is it, replace \(a=-1,b=1\) and get
\[\frac{1}{2}(e^{-x}(\sin(x)-\cos(x)))\]

Answer 5

or
\[\frac{1}{2}e^{-x}\sin(x)-\frac{1}{2}e^{-x}\cos(x)\] and then repeat the process

Answer 6

the fundamental formula is for \[\int\limits_{a}^{t}g(x)dx=g(t)\]

Answer 7

which prove by if f(x) is continuous on a closed interval[a,b] and if F(x) is any antiderivative of f(x) so that\[dF(x)/dx =f(x)\] then\[\int\limits_{a}^{b}f(x)dx=F(b)-F(a)\]

Answer 8

You can try with cos(x) ^= [e^(-ix)+e^(ix)]/2