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f''x = 2+cosx f(0)=-1 and f(pi/2)=0 Find f(x).
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find f(x)
integrate twice f'(x) = 2x +sin x + C f(x) = x^2 -cos x + Cx + K use the two initial values to find constants C,K
but both given values are to plug in the original function...how can we solve for Cx?
@dumbcow do u have any clue how to go further on?
first put x=0 in the equation f(x) = x^2 -cos x + Cx + K u'll get K
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but how do we know which one we have to use first?
if u put x=0 then c will be eliminated and u cal calculate K and then C but if u put x=pi/2 then both K and C will be there
f(0)=-1 -1+k=-1 k=0 now u have f(x)= x^2 -cos x + Cx now put x=pi/2 f(pi/2)=0 0=pi^2/4+(C*pi)/2 c=-(pi/2)
OMG i get it now!! thank uu
np:)
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