Thread: Getting Y in terms of x

1. Getting Y in terms of x

$

dx/dt=1/(t+1) dy/dt=2t
$

Write an equation expressing y in terms of x.

So far, I have $x=ln(t+1)+C$ and $y=t^2+C$

So,

t=e^(x-C)-1

and y would equal:

y=(e^(x-C)-1)^2+D

I don't know how to get rid of the 'Cs' or if the 'D' should be a C. I thought that there should be a D because the D represnts the constant from the y equation.
Not really sure if I'm doing this problem properly.

2. I would use the chain rule to get $\dfrac{dy}{dx}$ and find y from there

3. Is this what you mean?:

dy/dx=2t/(t+1)

Integrating gives,

y=2(t-ln(t+1))+C

so

y=2(e^(x-C)-1-x)+C

That still leaves me with C.

4. Originally Posted by JewelsofHearts
t=e^(x-C)-1

and y would equal:

y=(e^(x-C)-1)^2+D

I don't know how to get rid of the 'Cs' or if the 'D' should be a C. I thought that there should be a D because the D represnts the constant from the y equation.
You are correct. And C and D come from different integrations, so yet they would be different.

@ e^(i*pi)
$\displaystyle \frac{dy}{dx} = 2t^2 + 2t$

How do you get rid of the t?

-Dan