quick math question

[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

 

[RESmonkey]

C'mon, prove your egos.

 

[TheoPetro]

mmmmmm think harder you must


...and you thought this was going to be a helpful post

 

[Fenixgoon]

should be able to use an integrating factor.

 

[KLin]

Do your own @#$@ing homework. :roll:

 

[TheoPetro]

Originally posted by: KLin
Do your own @#$@ing homework. :roll:

I like this response

 

[bigal40]

Use implict differntiation.

also for the future the best place to get people to do your homework for you is cramster.com

 

[schneiderguy]

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

using the pythagorean theorem you get dy^6/dx^-1/4 = y^z + 55x - 12. factor the right side and you get (5y + a)^z. then use the associative property of subtraction and substitute and distribute into dy/dx=4 :thumbsup:

 

[RESmonkey]

But I'm integrating (bigal40).

 

[RESmonkey]

Originally posted by: schneiderguy

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

using the pythagorean theorem you get dy^6/dx^-1/4 = y^z + 55x - 12. factor the right side and you get (5y + a)^z. then use the associative property of subtraction and substitute and distribute into dy/dx=4 :thumbsup:

I have a feeling that you're messing with me...lol.

 

[RESmonkey]

???

 

[bonkers325]

you might want to post the full problem

 

[Toastedlightly]

well, does a separable differential equation mean anything to you?

 

[ManBearPig]

what are you asking exactly? you want the second derivative?

 

[Toastedlightly]

Alright... this is a separable diff eq. Do some cross multiplication.

dy = dx(y+2)

rearrange...

dy/(y+2) = dx

Integrate (remember constants and whatnot).

 

[RESmonkey]

I want to integrate that.

Like, integrate y + 2, only y is the function. (dy/dx = y + 2, integrate it, and then find the constant that allows y(o) = 2.

 

[RESmonkey]

Originally posted by: Toastedlightly
Alright... this is a separable diff eq. Do some cross multiplication.

dy = dx(y+2)

rearrange...

dy/(y+2) = dx

Integrate (remember constants and whatnot).

Can you elaborate on this? Sorry, but I'm just so confused.

 

[Toastedlightly]

Originally posted by: RESmonkey

Originally posted by: Toastedlightly
Alright... this is a separable diff eq. Do some cross multiplication.

dy = dx(y+2)

rearrange...

dy/(y+2) = dx

Integrate (remember constants and whatnot).

Can you elaborate on this? Sorry, but I'm just so confused.

What math class is this for? I'm just solving a differential equation. I don't want to give every thing away.

What I did was separate the variables. This is the method used to solve this diff eq. Cross multiplied the equation, then divided by (y+2).

 

[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)?

 

[RESmonkey]

The answer in the back is y = 4e^(x) - 2

 

[eLiu]

Originally posted by: Toastedlightly

Originally posted by: RESmonkey

Originally posted by: Toastedlightly
Alright... this is a separable diff eq. Do some cross multiplication.

dy = dx(y+2)

rearrange...

dy/(y+2) = dx

Integrate (remember constants and whatnot).

Can you elaborate on this? Sorry, but I'm just so confused.

What math class is this for? I'm just solving a differential equation. I don't want to give every thing away.

What I did was separate the variables. This is the method used to solve this diff eq. Cross multiplied the equation, then divided by (y+2).

OP: keep in mind here that separation of variables only applies for homogenous eqns & the special nonhomogenous case of a constant RHS.

In general for something like dy/dx = C*y(x) + f(x), you need to try something else... undetermined coefs will likely handle it easily. For example, if you had dy/dx = y + cos(x), separation would screw you.

Integrating factors are more for nonconstant coefficient nastiness.

 

[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+6) = 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+6). To start, one must know the rule that the integral of 1/x = natural log (ln) of x (ln(x)). Follow so far?

 

[RESmonkey]

I'm just gonna skip it. It's one of two problems I couldn't get in the review for a test tomorrow, oh well.

If you want the exact direction: Solve the initial value problem analytically. Support your solution by overlaying its graph on a slope field of the differential equation.

 

[RESmonkey]

yes, I follow.

 

[Toastedlightly]

Originally posted by: eLiu

Originally posted by: Toastedlightly

Originally posted by: RESmonkey

Originally posted by: Toastedlightly
Alright... this is a separable diff eq. Do some cross multiplication.

dy = dx(y+2)

rearrange...

dy/(y+2) = dx

Integrate (remember constants and whatnot).

Can you elaborate on this? Sorry, but I'm just so confused.

What math class is this for? I'm just solving a differential equation. I don't want to give every thing away.

What I did was separate the variables. This is the method used to solve this diff eq. Cross multiplied the equation, then divided by (y+2).

OP: keep in mind here that separation of variables only applies for homogenous eqns & the special nonhomogenous case of a constant RHS.

In general for something like dy/dx = C*y(x) + f(x), you need to try something else... undetermined coefs will likely handle it easily. For example, if you had dy/dx = y + cos(x), separation would screw you.

Integrating factors are more for nonconstant coefficient nastiness.

He's just in AP Calc BC... Calc II for the college bound. He doesn't need any of that nasty. I'm just taking Calc III, so Im learning that now.