First slide
Methods of integration
Question

If ∫eaxcosbx=e2x29f(x)+C and f′′(x)=kf(x) then k is equal to

Moderate
Solution

 We have, eaxcosbxdx=eaxa2+b2(acosbx+bsinbx)+Ce2x22+52f(x)+C=eaxa2+b2(acosbx+bsinbx)+Ca=2,b=5 and f(x)=acosbx+bsinbxf′′(x)=b2f(x)=25f(x)|k|=25

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