Afull tank initially (at t=0) contains 14 gallons. then water is removed at a rate of 1+t gallons per minute where t is the time in minutes. (a) how much water remains in the tank after t minutes?

Answer: the amount of water remaining after t minutes is      ======let y(t) represent number of gallons at a time t. then we would have y(0) = 14. this is an initial value problem with differential equations.the following differential equation can be created from the given information      we add a negative sign over the entire rate because the question states that the water is being removed. if we did not have the negative, the rate would be positive and thus increase y as time goes on.we solve this by treating dy/dt as a ratio of differentials and separate the variables. move dt to the right side    integrate both sides of the equation, making sure to leave a constant of integration.    using our initial condition, it says that when t = 0, y = 14. so solving for c:     thus y as a function of time can be written as    the amount of water remaining after t minutes is -t - t²/2 + 14.