Imade this the highest point value, and i will give brainliest to whoever does it well and right!   which of the following differential equations with initial condition yields a solution curve that exhibits modified exponential growth such that y goes to negative infinity as t goes to positive infinity? (hint: check each one by solving the differential equation quickly. remember, to solve the differential equation the quick and easy way, recall the fact that the solution to the differential equation \frac{dy}{dt} = k(y-a) is y = a e^{kt} +a. so you just need to write the differential equation in the proper form first. then, of course, you must solve for the constant a using the initial condition.) a.) \frac{dy}{dt} = -2(y-200) with initial condition y(0) = 100 b.) \frac{dy}{dt} = -3.5(y+123) with initial condition y(0) = 100 c.) \frac{dy}{dt} = 3.5(y+123) with initial condition y(0) = 50 d.) \frac{dy}{dt} = 2(y-200) with initial condition y(0) = 50 e.) \frac{dy}{dt} = 2(y+200) with initial condition y(0) = 50

The answer to this question is (d). with initial condition y(0) = 50======if we use the hint, we convert these all of these differential equations like so    so we will check each one, and then use the initial condition to solve for the constant a. we will determine if any one of the equations results in an exponential function with a behavior that is described by, "y goes to negative infinity as t goes to positive infinity" (i.e.  ).=======checking (a)    solving for a using initial condition y(0) = 100 (when t = 0, y = 100)    the solution to this differential equation is therefore  . as t approaches infinity:     we used the fact that  . (e raised to a very large negative exponent approaches 0.)so this is not the answer.======checking (b)    solving for a using initial condition y(0) = 100 (when t = 0, y = 100)    the solution to this differential equation is therefore  . the limit of the solution as t approaches infinity is    this is not the answer.======checking (c)    solving for a using initial condition y(0) = 50 (when t = 0, y = 50)    therefore, the solution to this differential equation is  . the limit of this solution as t approaches infinity is    since  this is positive infinity. this is not the answer.======checking (d)    solving for a using initial condition y(0) = 50  (when t = 0, y = 50)    therefore, the solution to this differential equation is  . the limit of this solution as t approaches infinity is    the answer to this question is (d).if we did the same thing for (e), we would end up with a = 250, which wouldn't result in a +∞ limit

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.