Question

dx/dt=9-4x^2, x(0)=0
Seperate variables use partial fractions to solve the initial value

Answer 1

hmm..maybe\[\frac{dx}{dt} =9-4x^2 => \frac{dt}{dx} =\frac{1}{9-4x^{2}} dx => t=\int\limits\frac{1}{9-4x^{2}} dx\]not sure tho..havent done these in a while.

Answer 2

yes set up is
\[\int\limits_{?}^{?}9-4x^2= \int\limits_{?}^{?}dt\] 9-4x^2 dx=\int\limits_{?}^{?} dt\]

Answer 3

\[x=\int\limits9-4x^{2} dx\]??
doesnt look right tho

Answer 4

right side is t + C
but left side i'm lost

Answer 5

sorry left side woud be
interval 1/(9-4x^2) dx

Answer 6

so what i had; is right?

Answer 7

yes

Answer 8

ok..use trig sub

Answer 9

help

Answer 10

\[x=\frac{9}{4} \sin\theta \]
for trig sub

or you can use table of standard integrals

Answer 11

need help with ∫9−4x 2 dx do i use u=9-4X^2 so du= -8x
but them what?

Answer 12

I would separate the variables like so:
\[ \frac{dx}{dt} = 9 - 4x^2 \\
\frac{1}{9 - 4x^2} \frac{dx}{dt} = 1 \\
\frac{1}{9 - 4x^2} dx = dt \\
\int \frac{1}{9 - 4x^2} dx = \int 1 dt \\
\int \frac{1}{9 - 4x^2} dx = t + C \]
\[ \int \frac{1}{9 - 4x^2} dx \]
Then, use partial fractions on this integral here.

Answer 13

you can use trig-sub; partial fractions is too tedious.

Answer 14

Well, that may be true, but the original post seems to ask specifically for partial fractions...
```
Seperate variables use partial fractions to solve the initial value
```

Answer 15

i need help with the partial fraction of 1/(9-4x^2)
can you show me the steps, please, :-)

Answer 16

Well, the first step here is to identify the factors of the denominator. Do you know how to factor this denominator?

Answer 17

yes 3-2x and 3+2x

Answer 18

Okay, good. :)
We can rewrite this integral a bit then:
\[ \int \frac{1}{(3 - 2x)(3 + 2x)} dx \]
Partial fractions is basically the process of writing this fraction in terms of a sum of, usually, more simple fractions, like this: There are some differences depending on how the denominator is factored, but here is the specific way for our denominator: non-repeated linear factors only:
\[ \frac{1}{(3 - 2x)(3 + 2x)} = \frac{A}{3 - 2x} + \frac{B}{3 + 2x} \]
We just have to find the A and B here.

Answer 19

ok let me try

Answer 20

I recall there are multiple ways to do this, so I will be using one specific technique that I am familiar with. (It may not be the same as what you use)

Multiplying both sides by the denominator of the left-hand side, it becomes:
\[ \frac{\cancel{(3 - 2x)(3 + 2x)}}{\cancel{(3 - 2x)(3 + 2x)}} = \frac{A\cancel{(3 - 2x})(3 + 2x)}{\cancel{3 - 2x}} + \frac{B(3 - 2x)\cancel{(3 + 2x)}}{\cancel{3 + 2x}} \\
1 = A(3 + 2x) + B(3 - 2x)\\
1 = 3A + 2Ax + 3B - 2Bx \\
1 = (2A - 2B)x + (3A + 3B)\]
Basically, we can equate the coefficients here.
\(2A - 2B = 0\)
\(3A + 3B = 1\)
This is a system of linear equations, which may be solved for the A and B.

Answer 21

* If it is unclear where the \(2A - 2B = 0\) part comes from, remember that \(1 = 0x + 1\), and we equate the (2A + 2B) coefficient on the right side to this 0.

Answer 22

the book says
1/6(3-2x)+1/6(3+2x)
i dont get it

Answer 23

thats the partial fraction of
1 (3−2x)(3+2x)

Answer 24

Have you read my above explanation or do not understand it?

Answer 25

do not understand it because its been 22 years since I was in college and Im a litttle lost right now.

Answer 26

Sorry, was trying to make this:
http://i49.tinypic.com/2hfigr8.gif
Maybe this is easier to follow? It goes through the basic process to get the system, putting up the next step every 2 seconds. It stops at the end.

Answer 27

i follow this p://i49.tinypic.com/2hfigr8.gif
find but that still doesnt get me to
1/(6(3-2x))+1/(6(3+2x))

Answer 28

Okay, now we'll solve that linear system, which is just Algebra work now:
2A - 2B = 0 Multiply both sides by 3
3A + 3B = 1 Multiply both sides by 2

6A - 6B = 0
6A + 6B = 2 Add together to eliminate B:

12A = 2
A = 1/6

6(1/6) - 6B = 0
1 - 6B = 0
1 = 6B
1/6 = B

We can put these in for A and B to get what you posted.
\[
\frac{1}{(3 - 2x)(3 + 2x)} = \frac{1 \over 6}{3 - 2x} + \frac{1 \over 6}{3 + 2x} \\
\frac{1}{(3 - 2x)(3 + 2x)} = \frac{1}{6}\frac{1}{3 - 2x}\ + \frac{1}{6} \frac{1}{3x + 2}] \]

Answer 29

Now, this allows us to rewrite that integrand into a sum of two fractions. I'll factor out the 1/6 and we get:
\[ \int \frac{1}{9 - 4x^2} dx = \int \frac{1}{6} \left( \frac{1}{3 - 2x} + \frac{1}{3 + 2x} \right) dx \]

Answer 30

Sorry, it's kind of late, I'll try to hurry up:
We may bring the 1/6 out front and integrate each term:
\[ \frac{1}{6} ( \int \frac{1}{3-2x} dx + \int \frac{1}{3 + 2x} dx ) \]
From this, we can make some u-substitutions and get a form much more like:
\[ \int \frac{1}{u} du \]
Which is just \(\ln |u|\)

Answer 31

\[ \int \frac{1}{3 - 2x} dx\]
Let \(u = 3 - 2x\)
\(du = -2 dx \implies \neg \frac{1}{2} du = dx \)
\[ \int \frac{1}{u} (\neg \frac{1}{2} du) \\
\neg \frac{1}{2} \int \frac{1}{u} du \\
\neg \frac{1}{2} \ln |u| \\
\neg \frac{1}{2} \ln |3 - 2x|
\]
I assume you are familiar with this process?

Answer 32

ok i get this now... so the I get
-1/12 ln (3-2x)+1/12 ln(3+2x)=t
solve for x
need help with this part.
and yes I familiar with u sub.

Answer 33

Are you required to solve for x? Seems pretty difficult to solve this without maybe a hyperbolic trig function, and would be easier to leave in implicit form...

Answer 34

maybe not i'll fid out.. thanks for your help. you were very helpful

Answer 35

The immediate process to solve is just to take x = 0, t = 0 and solve for the constant C:
-1/12 ln (3-2x)+1/12 ln(3+2x)=t + C
> The + C is from indefinite integration.

Answer 36

Oh, I got it. You need a bit of manipulation to do it:
We could use a logarithm property to get one logarithm
log(a) + log(b) = log(ab)
log(a) - log(b) = log(a/b)


-1/12 ln (3-2x)+1/12 ln(3+2x)=t + C (don't forget that +C since its indefinite integral) factor out a 1/12

1/12(-ln(3 - 2x) + ln(3 + 2x)) = t + C reordering
1/12(ln(3 + 2x) - ln(3 - 2x)) = t + C use log property
1/12 ln( (3+ 2x)/(3 - 2x) ) = t + C multiply both sides by 12
ln( (3 + 2x)/(3 - 2x) ) = 12t + C raise both sides as a power of e
(3 + 2x)/(3 - 2x) = e^(12t+C) Multiply (3 - 2x) denominator to both sides
3 + 2x = (3 - 2x) e^(12t + C) Distribute
3 + 2x = 3 e^(12t + C) - 2x e^(12t + C) Get terms with x on one side
2x + 2xe^(12t + C) = e^(12t + C) - 3 Factor out the x.
x(2 + 2e^(12t + C) = e^(12t + C) - 3 Divide by (2 + 2e^(12t + C))
x = (e^(12t + C) - 3) / (2e^(12t + C) + 2)

Answer 37

In that form, C is a bit tougher to solve for though for the initial value problem. :P

Answer 38

\[x=\frac{ -2 }{ 4 }k e^{t}\] is this the correct answer for \[\frac{ dx }{ dt}=4x-2\]

Answer 39

\[\frac{ dy }{ dt}=-3y-2 ; y(2) = 0 .. \] when i am solving this equation i have got \[y=\frac{ -2 }{ 3 }+Ke ^{-3t}\] .. now please help me for next step with y(2) = 0...