Find differential equation y ′ = f ( t , y ) satisfied by y...

Jaydan Aguirre

Jaydan Aguirre

Answered

2022-07-09

Find differential equation y = f ( t , y ) satisfied by y ( t ) = 4 e 2 t + 3
Solution:
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 6
So 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 6
and 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.

Answer & Explanation

behk0

behk0

Expert

2022-07-10Added 14 answers

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.
Issues
How 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 .

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