Answer :

Formula:-


(i) if a differential equation is ,


then y(I.F) = ∫Q.(I.F)dx + c, where I.F = e∫Pdx


(ii) ∫dx = x + c



This is a linear differential equation, comparing it with



P = 1, Q = e–2x


I.F = e∫Pdx


= e∫Pdx


= ex


Solution of the equation is given by


y(I.F) = ∫Q.(I.F)dx + c


y(ex) = ∫e–2x ex dx + c


y(ex) = ∫e–x dx + c


y(ex) = –e–2x + c


y = –e–2x + c e–x


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