in Calculus edited by
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The solution of the differential equation (d y /dx)=ky , Y(0)= C is

(a) x =ce-xy

(b) y= cekx

(c) x =ke cy

(d) y= ce-kx
in Calculus edited by
by
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1 Answer

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(1/y) dy = kdx Now integrate the equation which gives

log y = kx + c'

log y = kx + log c ( c'= log c)

y = e(kx +log c)  = cekx (option B)

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