Question

URGENT!! HELP ME PLEASE!

given dy/dx = a(e^-x)+2 and that when x=0, dy/dx =5 and y=1, then when x=2, what is y equals to?

Answer 1

Do you know how to anti-differentiate? Or are expected to solve this using a different method?

Answer 2

you have to use anti diff.
i know how to anti differentiate but i cant get the answer :(

Answer 3

Let's start by solving for the missing A value c:

Plugging in the given values will give us,\[\large \frac{dy}{dx}=ae^{-x}+2 \qquad \rightarrow \qquad 5=ae^0+2 \qquad \rightarrow \qquad 3=a\]Were you able to get that part? c:

Answer 4

yup!!

Answer 5

\[\large \frac{dy}{dx}=3e^{-x}+2\]We'll rewrite the dx on the other side,\[\large dy=3e^{-x}+2 \; dx\]

Integrating gives us,\[\large \int\limits dy=\int\limits 3e^{-x}+2 \; dx \qquad \rightarrow \qquad y=-3e^{-x}+2x+C\]

Answer 6

there i found that i cant get C :(

Answer 7

Any confusion? If you're just being introduced to "anti-differentiation", then the integral sign might be confusing.

I forgot about that maybe :D

Answer 8

Ok let's seeeeee

Answer 9

So they gave us an initial condition to use.
When x=0, y=1.

Plugging those in gives us,\[\large 1=-3e^0+2\cdot 0 + C\]

Hmm I'm not sure why they gave us an initial condition containing information for the derivative and anti-derivative.
That's kinda strange...

I mean I guess it was so we could find A, then C later on.
It's just weird :3 Hopefully I didn't make a silly error in there anywhere.

Answer 10

it is 8 :(

Answer 11

err..

Answer 12

Oh that 2 was being multiplied by 0, woops my bad.

So C=4 I think? :o

Answer 13

We'll plug our new-found C value into our equation,\[\large y=-3e^{-x}+2x+4\]

Answer 14

They give us another x value, and ask us to find the corresponding y value.

So when x=2,\[\large y=-3e^{-2}+2\cdot 2+4\]