Q. 2 A

# For each of the e

It is given that xy = a ex + b e–x + x2

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 = = x(aex +be-x + 2) + 2(aex - be-x + 2) –x(aex +be-x + x2) + x2 – 2

= (axex +bxe-x + 2x) + 2(aex - be-x + 2) –x(aex +be-x + x2) + x2 – 2

= 2aex -2be-x +x2 + 6x -2

≠ 0

LHS ≠ RHS.

Therefore, the given function is not the solution of the corresponding differential equation.

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