Determine whether the given function is a solution to the given differential equation. d'e de -e-+ 30 = - 39 e 5t dt 0= e21-4 e 5t dt de -0 + 30 = - 39 e 5t because when e 21 -4 e 5t is substituted for 0. is substituted for and lis substituted for dt? d0 The function 0- -4 e 5t a solution to the differential equation the two sides of the differential equation dt equivalent on any intervals of L are are not

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