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Question

If y=eax. cos bx, then prove that d2ydx22adydx+(a2+b2)y=0


Solution

y=eax.cosbx

dydx=aeax.cos bxbeax.sin bx...........(i)dydx=aybeax.sin bxd2ydx2=adydxb(aeax.sin bx+beax.cos bx)d2ydx2=adydxbaeax.sin bxb2eax.cos bxd2ydx2=adydxa(aydydx)b2y

[Substituting beax sin bx from (i)]

d2ydx2=adydxa2y+adydxb2yd2ydx22adydx+(a2+b2)y=0

Hence Proved

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