Answer :

It is given that y = ex(acosx + bsinx) = aexcosx + bexsinx


Now, differentiating both sides w.r.t. x, we get,





Now, again differentiating both sides w.r.t. x, we get,






Now, Substituting the values of ’ and in the given differential equations, we get,


LHS =


=2ex(bcosx – asinx) -2ex[(a + b)cosx + (b –a ) sinx] + 2ex(acosx + bsinx)


=ex[(2bcosx – 2asinx) - (2acosx + 2bcosx) - (2bsinx – 2asinx) + (2acosx + 2bsinx)]


= ex[(2b – 2a – 2b + 2a)cosx] + ex[(-2a – 2b + 2a + 2bsinx]


= 0 = RHS.


Therefore, the given function is the solution of the corresponding differential equation.


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