# Thread: Find the value of y when x=2.

1. ## Find the value of y when x=2.

Given that dy/dx=a*e^(-x)+1 and that when x=0, dy/dx=3 and y=5, find the value of y when x=2.
So I subbed in x=0 and dy/dx=3 into the derivative to find a, which ended up being a=2.
Then I anti differentiated dy/dx to get y=a*e^(-x)+x.
I subbed in y=5 and x=0 into that equation to find a, which was a=-5.
And then I didn't really know what to do from there.
I don't even know if what I did was necessary to answer the question either.

2. ## Re: Find the value of y when x=2.

Originally Posted by juliacoolness
Given that dy/dx=a*e^(-x)+1 and that when x=0, dy/dx=3 and y=5, find the value of y when x=2.
So I subbed in x=0 and dy/dx=3 into the derivative to find a, which ended up being a=2.
Then I anti differentiated dy/dx to get y=a*e^(-x)+x.
I subbed in y=5 and x=0 into that equation to find a, which was a=-5.
And then I didn't really know what to do from there.
I don't even know if what I did was necessary to answer the question either.
1. You made a tiny mistake by calculating y:

$\dfrac{dy}{dx}=2 \cdot e^{-x} +1 ~\implies~y=-2 \cdot e^{-x}+x+c$

2. Now you know that f(0) = 5. That means:

$f(0)=-2+c=5~\implies~c=7$

3. So the complete equation of the function is:

$y = f(x) = -2 \cdot e^{-x}+x+7$

4. Now plug in x = 2.

3. ## Re: Find the value of y when x=2.

Thanks so much mate.
I'm so glad you pointed that out to me.