quick math question

[Toastedlightly]

Originally posted by: Toastedlightly

Originally posted by: RESmonkey
AP Calc BC.

I think my teacher did this, but got side tracked and never solved it. I have unfinished notes somewhere.

How do u integrate it when dy is being dividded by that? Do you just worry about the right side (dx)?

If you have a book, look up separable differential equations, or most likely just differential equations. I can work through the problem, but I suck at teaching.

From where I left off, we have dy/(y+2) = dx
Now we integrate both sides (each w/ respect to their own variable, either y or x).

We know the integral of dx is x. This is easy. The part you seem to be having problems with is the dy/(y+2). To start, one must know the rule that the integral of 1/x = natural log (ln) of x (ln(x)). Follow so far?

Now, we have to take the integral of 1/(y+2) * dy by using substitution, using the fact that we know what the integral of 1/x equals.

EDIT: wow, did I mess up the equation?

 

[RESmonkey]

Got it. It's y + 2, btw. I have ln(y+2) + C = x + C. We're integrating on both sides, right?

 

[hypn0tik]

Originally posted by: RESmonkey
dy/dx = y + 2 , y(o) = 2

How do I integrate dy/dx when y is inside the derivative? I'm thinking it has something to do with e^x, since it's derivative is itself.

Thanks

You can see right away that it's a separable equation. Put all the y's on one side and all the x's on the others.

In this case:

1/(y + 2) dy = dx
Integrate both sides
For the integral on the right, it should be trivial.

For the integral on the left, you can use a simple substitution if it'll help you visualize it easier.

Let u = y + 2
du = dy

--> 1/(y + 2) dy = 1/u du

Integrating, you get:

ln |u| = x + C
ln |y + 2| = x + C

--> e^(x + C) = y + 2
e^C*e^x = y + 2

Since e^C is just a constant, you can call it D
D*e^x = y + 2

Now, you have your initial condition: y(0) = 2

D*e^0 = y + 2
y = D - 2 = 2
--> D = 4

Therefore, the solution is: y = 4e^x - 2

Assuming of course, I haven't made any silly mistakes...


Edit You posted while I posted. You only need the constant on one side of the equation after you integrate. If you have it on both sides, you can just have one constant absorb the other.

 

[Toastedlightly]

Originally posted by: RESmonkey
Got it. It's y + 2, btw. I have ln(y+2) + C = x + C. We're integrating on both sides, right?

yup. from there, you should be able to figure it out. Sorry about that, I'm actually doing some vibrating spring problems right now and I got my variables screwed up.

 

[Toastedlightly]

Originally posted by: hypn0tik

Originally posted by: RESmonkey
dy/dx = y + 2 , y(o) = 2

How do I integrate dy/dx when y is inside the derivative? I'm thinking it has something to do with e^x, since it's derivative is itself.

Thanks

You can see right away that it's a separable equation. Put all the y's on one side and all the x's on the others.

In this case:

1/(y + 2) dy = dx
Integrate both sides
For the integral on the right, it should be trivial.

For the integral on the left, you can use a simple substitution if it'll help you visualize it easier.

Let u = y + 2
du = dy

--> 1/(y + 2) dy = 1/u du

Integrating, you get:

ln |u| = x + C
ln |y + 2| = x + C

--> e^(x + C) = y + 2
e^C*e^x = y + 2

Since e^C is just a constant, you can call it D
D*e^x = y + 2

Now, you have your initial condition: y(0) = 2

D*e^0 = y + 2
y = D - 2 = 2
--> D = 4

Therefore, the solution is: y = 4e^x - 2

Assuming of course, I haven't made any silly mistakes...

Can't just give away the answer. Make 'im work for it!

 

[RESmonkey]

Thanks guys ! I finally get it...

I shall slaughter that test.

 

[RESmonkey]

I worked it out looking off of it. I understood it, though, lol.

 

[DrPizza]

Good luck on your test.

<snickers about a few of the responses in the thread>

Just remember, it's simply a letter. It doesn't matter what the variable is.

integral of dy/(y+2) is no different than the integral of dx/(x+2)