Asked by Rajan
f''(x)= 4e^x-9sin(x) f(0)=3 f(pi/2)=0
I am to find f(x); but I am confused about how to do that. Could I please get some help? I think I have to find f'(x) and then go backwards from there, but my answer doesn't seem to work.
I am to find f(x); but I am confused about how to do that. Could I please get some help? I think I have to find f'(x) and then go backwards from there, but my answer doesn't seem to work.
Answers
Answered by
oobleck
too bad you didn't bother to show your work.
f" = 4e^x - 9sinx
f' = 4e^x + 9cosx + c1
f = 4e^x + 9sinx + c1*x + c2
using the two values of f(x), we have
4+0+0+c2 = 3 ==> c2 = -1
4e^(π/2) + 9 + c1 * π/2 -1 = 0
c1 = -(8+4e^(π/2))/(π/2) = -8/π (2+e^(π/2))
f(x) = 4e^x + 9sinx - 8/π (2+e^(π/2)) x - 1
f" = 4e^x - 9sinx
f' = 4e^x + 9cosx + c1
f = 4e^x + 9sinx + c1*x + c2
using the two values of f(x), we have
4+0+0+c2 = 3 ==> c2 = -1
4e^(π/2) + 9 + c1 * π/2 -1 = 0
c1 = -(8+4e^(π/2))/(π/2) = -8/π (2+e^(π/2))
f(x) = 4e^x + 9sinx - 8/π (2+e^(π/2)) x - 1
Answered by
Rajan
Ah, thank you! My apologies for not showing my work; I got f(x)= 4e^x+9sinx+cx but forgot that it would be cx+c2 so I was just confused on how to solve it with the two f(x) values given. Thanks again!