Jaydan Aguirre

Answered

2022-07-09

Find differential equation ${y}^{\mathrm{\prime }}=f\left(t,y\right)$ satisfied by $y\left(t\right)=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}+3$
Solution:
Compute derivative of y,
${y}^{\mathrm{\prime }}=8\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$
Write right hand side above, in terms of the original function y, that is,
$y-3=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$
Get a differential equation satisfied by y, namely
${y}^{\mathrm{\prime }}=2y-6$
So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of $2y-6$ you should end up with the original $y\left(t\right)=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}+3$???
It seems that there two different derivatives of y(t)
one is:
${y}^{\mathrm{\prime }}=8\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$
the other is:
${y}^{\mathrm{\prime }}=2y-6$
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

Expert

2022-07-10Added 14 answers

Compute the square of y,
${y}^{2}=4+2\sqrt{3}.$
Write the right-hand side above in terms of the original number y. That is,
${y}^{2}=2y+2.$
Rearrange to get a polynomial equation satisfied by y, namely
${y}^{2}-2y-2=0.$
Issues
How is this a solution?
This is a solution because, if you plug $y=1+\sqrt{3}$ the polynomial $p\left(x\right)={x}^{2}-2x-2$, you get zero. In other words, you found a polynomial p that makes the equation $p\left(y\right)=0$ true.
Does it mean that if you somehow take a square root of $2y+2$, you should end up with the original $y=1+\sqrt{3}$?
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 ${y}^{2}=4+2\sqrt{3}$. The other is ${y}^{2}=2y+2$.
For most numbers x, the numbers ${x}^{2}$ and $2x+2$ would indeed be different. The fact that ${y}^{2}$ and $2y+2$ are the same is a special property of the number $y=1+\sqrt{3}$.

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