Simple physics problem using differential equation modelingThere are 50 litres of dilution, 90% water and 10% of alcohol in a tank. Then, 50% of water and 50% of alcohol is poured in a tank, at the speed of 5 litres per minute, and the dilution is outflowing from a tank at the speed of 5 litres per minute. How much alcohol is in a tank after 10 minutes?I have a solution that use differential equations, but I can't understand how to model the problem by differential equations. Obtained differential equation is: 2 − 5 50 − t y ( t ) = d y d t ⇒ y ′ + 5 50 − t y = − 2which is a linear differential equation.General solution: y ( t ) = C ( 50 − t ) 5 + 1 2 ( 50 − t )Now, setting y 0 = 5, which is the amount of alcohol at the beginning, we evaluate constant C ⇒ C = − 20 50 5 Going back to the general solution: C = − 20 50 5 litres.Also, what would be the approach without using differential equations?

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Simple physics problem using differential equation modelingThere are 50 litres of dilution, 90% water and 10% of alcohol in a tank. Then, 50% of water and 50% of alcohol is poured in a tank, at the speed of 5 litres per minute, and the dilution is outflowing from a tank at the speed of 5 litres per minute. How much alcohol is in a tank after 10 minutes?I have a solution that use differential equations, but I can&#039;t understand how to model the problem by differential equations. Obtained differential equation is:  2  −      5          50      −      t        y  (  t  )  =            d      y              d      t        ⇒      y    ′    +      5          50      −      t        y  =  −  2which is a linear differential equation.General solution:  y  (  t  )  =  C  (  50  −  t      )    5    +      1    2    (  50  −  t  )Now, setting       y    0    =  5, which is the amount of alcohol at the beginning, we evaluate constant   C  ⇒  C  =  −      20          50      5      Going back to the general solution:  C  =  −      20          50      5       litres.Also, what would be the approach without using differential equations?

Quinten Cervantes · Accepted Answer

Step 1This problem is a bit of a special (easy) case, because the amount of solution in the tank does not change. If we assume perfect instantaneous mixing, then the two variables corresponding to the amounts of water and alcohol in the tank are decoupled and we get            d      y              d      t        =  −  5        y    50    +      5    2      .The first term on the right-hand side accounts for the flow out of the tank, and says that the rate at which alcohol is going out at time t is proportional to the fraction of alcohol in the solution in the tank at time t.Step 2The second term is the rate at which alcohol is coming in (half of the 5 liters per minute).This is one of the simplest differential equations to be solved. The general solution to the homogeneous equation is   C      e          −      t              /            10      . The particular solution has the form   A  t  +  B. Substitution gives   A  =  0 and   B  =  25. Finally, imposing the initial condition   y  (  0  )  =  5, we get   5  =  C  +  25, or   C  =  −  20.All in all, taking the initial condition into account, we get  y  (  t  )  =  25  −  20        e          −      t              /            10          .We can check that   y  (  0  )  =  5 and       lim          t      →      ∞        y  (  t  )  =  25, in perfect agreement with intuition.

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