Q. 2 B3.6( 14 Votes )

# For each of the e

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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