questions are below:

Determine whether the given function is a solution to the given differential equation. 0 =5 e 5t - 2 3t d'e de + 40= - 3e 3t dt 2 The function 0 = 5 e 5t - 2 e 3t AP d20 a solution to the differential equation at + 40 = - 3 es, because when 5 e 5t - 2 e 3t is substituted for 0, |is substituted for f and |is substituted for- 2 . the two sides of the differential equation equivalent on any intervals of t.Determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship does define y implicitly as a function of x and use implicit differentiation. e 4xy + 2y? = 2x2+3, dy xe - 4xy - y - = dx ye -4xy + X Applying implicit differentiation to the equation gives , which dy xe - 4xy - y equivalent to Therefore, e 4xy + 2y? = 2x2 + 3 a solution to the differential equation. dx ye - 4xy +XSolve the initial value problem. 1 - yexy - - dx + xexy X + 2 dy = 0, y(2) = 3 The solution is (Type an equation using x and y as the variables. Type an implicit solution.)