If ∫eaxcosbx=e2x29f(x)+C and f′′(x)=kf(x) then k is equal to
We have, ∫eaxcosbxdx=eaxa2+b2(acosbx+bsinbx)+C⇒e2x22+52f(x)+C=eaxa2+b2(acosbx+bsinbx)+C⇒a=2,b=5 and f(x)=acosbx+bsinbx⇒f′′(x)=−b2f(x)=−25f(x)∴|k|=25