Simple separation of variables, why is this not a solution?

[medicalphysicsguy]

Greetings,

Chapter 1, my DQ textbook asks us to evaluate potential solutions to:

\(\displaystyle \frac{dy}{dt}=\frac{y+1}{t+1}\)

It says

\(\displaystyle y = t^2-2\)

is not a solution.

But I get:

\(\displaystyle \frac{t^2-1}{t+1} = \frac{(t+1)(t-1)}{t+1} = t-1\)

therefore

\(\displaystyle dy = (t-1)dt\)

or

\(\displaystyle y = \frac{1}{2}t^2 - t\)

Can anyone tell me what is wrong with my math here?

Thanks,
Eric

 

[Deleted member 4993]

Greetings,

Chapter 1, my DQ textbook asks us to evaluate potential solutions to:

\(\displaystyle \frac{dy}{dt}=\frac{y+1}{t+1}\)

It says

\(\displaystyle y = t^2-2\)

is not a solution.

But I get:

\(\displaystyle \frac{t^2-1}{t+1} = \frac{(t+1)(t-1)}{t+1} = t-1\)

therefore

\(\displaystyle dy = (t-1)dt\)

or

\(\displaystyle y = \frac{1}{2}t^2 - t\)

Can anyone tell me what is wrong with my math here?

Thanks,
Eric

Nothing wrong ... it shows that y = t² - 2 is not a solution

You started with y = t² - t and you ended up with y = 1/2 t² - t

It is like starting with x =2 but ending up with x =1. That only proves that x is not equal to 2.

 

[medicalphysicsguy]

Makes sense now, thanks.