Find differential equation y ′ = f ( t , y ) satisfied by y ( t ) = 4 e 2 t + 3Solution:Compute derivative of y, y ′ = 8 e 2 t Write right hand side above, in terms of the original function y, that is, y − 3 = 4 e 2 t Get a differential equation satisfied by y, namely y ′ = 2 y − 6So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of 2 y − 6 you should end up with the original y ( t ) = 4 e 2 t + 3???It seems that there two different derivatives of y(t)one is: y ′ = 8 e 2 t the other is: y ′ = 2 y − 6and I don't get it, can someone explain?Also a bit offtopic, but the way y'=f(t,y) is written kinda bugs me.Shouldn't it be written like y'=f(t,y(t)) to show that the function f contains t as an independent variable and the function y(t) which contains variable t as an input to itself (dependent variable t) ??? That's kinda an essential information, so surprised it's omitted in the writings.

Question

Find differential equation       y          ′        =  f  (  t  ,  y  ) satisfied by   y  (  t  )  =  4        e          2      t        +  3Solution:Compute derivative of y,      y          ′        =  8        e          2      t      Write right hand side above, in terms of the original function y, that is,  y  −  3  =  4        e          2      t      Get a differential equation satisfied by y, namely      y          ′        =  2  y  −  6So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of   2  y  −  6 you should end up with the original   y  (  t  )  =  4        e          2      t        +  3???It seems that there two different derivatives of y(t)one is:      y          ′        =  8        e          2      t      the other is:      y          ′        =  2  y  −  6and I don&#039;t get it, can someone explain?Also a bit offtopic, but the way y&#039;=f(t,y) is written kinda bugs me.Shouldn&#039;t it be written like y&#039;=f(t,y(t)) to show that the function f contains t as an independent variable and the function y(t) which contains variable t as an input to itself (dependent variable t) ??? That&#039;s kinda an essential information, so surprised it&#039;s omitted in the writings.

behk0 · Accepted Answer

Compute the square of y,      y    2    =  4  +  2      3    .Write the right-hand side above in terms of the original number y. That is,      y    2    =  2  y  +  2.Rearrange to get a polynomial equation satisfied by y, namely      y    2    −  2  y  −  2  =  0.IssuesHow is this a solution?This is a solution because, if you plug   y  =  1  +      3    the polynomial   p  (  x  )  =      x    2    −  2  x  −  2, you get zero. In other words, you found a polynomial p that makes the equation   p  (  y  )  =  0 true.Does it mean that if you somehow take a square root of   2  y  +  2, you should end up with the original   y  =  1  +      3  ?Yes! This is a surprising fact, one we might not have known before solving the problem. You can check it with a calculator.It seems there are two different squares of y. One is       y    2    =  4  +  2      3  . The other is       y    2    =  2  y  +  2.For most numbers x, the numbers       x    2   and   2  x  +  2 would indeed be different. The fact that       y    2   and   2  y  +  2 are the same is a special property of the number   y  =  1  +      3  .