# Math

Can you help me with this? I did this many times and got different answers each time.

Find f.
f ''(x) = 3e^x + 3sin(x)
f(0) = 0
f(π) = 0

My work:
f ''(x) = 3e^x + 3sin(t)
f'(x) = 3e^x - 3cos(t) + C
f(x) = 3e^x -3sin(t) + Cx + D

0=f(0)= 3e^0 - 3sin(0) + C(0) + D
D=-3

0=f(π)= 3e^π - 3sin(π) + Cπ -3
C= (3-3e^π)/π

F(x) = 3e^x -3sin(x) + (3-3e^π)/π x - 3

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1. f(0) = 0
f(π) = 0
Check if one of them is actually f'().

Assuming it's correct as above:
from:
f(x) =3e^x -3sin(x) + (3-3e^π)x/π - 3
We get:
f"(x)=3sin(x)+3e^x
f(0)=0
f(π)=0
which clearly satisfy all the given requirements. However, since there were two initial conditions for f(x), it is possible to have multiple solutions.

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