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HELP ASAP. Only Where A2,b-1 and C0 Are Different, Can You Solve the Differential Equation (dy/dt/ax

You need to think a bit before you start panicking. It would help.

Put those values in:

dx/dt ax by

dy/dt cx dy

where a 2, b -1, c 0, d 0

becomes:

dx/dt 2x - y

dy/dt 0

Now I really hope you can solve the equation dy/dt 0. From there it's not hard to solve for x:

dx/dt - 2x -y -k (where y k is a constant)

Multiply by the integrating factor e^(-2t):

e^(-2t) dx/dt - 2x e^(-2t) -k e^(-2t)

> d/dt (x e^(-2t)) -k e^(-2t)

> x e^(-2t) -k e^(-2t) / (-2) c

> x k/2 ce^(2t).

So the solution is x(t) k/2 ce^(2t), y(t) k for arbitrary constants c and k. You can check that this works: dy/dt 0, of course, and dx/dt 2c e^(2t) 2x - k 2x - y.

â€¢ Related Questions

How to prove that f(x) cx for some constant c that is an element of real numbers?

This is Cauchys functional equation. We will prove that

f(x)cx for all real x.

Proof: It is easy to show for positive integers n,

f(nx) nf(x)

which can be established by induction. Note that setting x1 and nk gives f(k)kf(1) for all positive integers k.

To prove for positive rationals, let xm/n. Then nxm and f(nx)f(m). So,

f(nx) nf(x) f(m) mf(1)

Thus,

f(m/n) f(x) (m/n)f(1)

which shows that f(x) cx for all positive rationals, x.

For negative rationals and x0 it is obvious that,

f(x) f(-x) f(x (-x)) f(0) 0 f(-x) -f(x) -cx f(x) cx for all rational x.

To complete the proof, we note that if f:RR and g: RR are continuous such that f(q)g(q) for rational q, then f(x) g(x) for all real numbers x. This follows from taking limits on f(x) and g(x) as they are both continuous.

Q.E.D

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What is Orthogonal trajectories in Differential equation is x^(2) y Cx - y?

The differential equation which generates the family of curves

xy Cx - y

can be found solving the previous wrt C

C y(x 1)/x

and differentiating both sides wrt x

xy' (x 1) 2xy - y(x 1)/x 0

xy'(x 1) 2xy - xy - y/x 0

xy'(x 1) 2xy - xy - y 0

solve wrt y'

y' y(1 - x)/(x(x 1))

and then substitute y' with - 1/y' because tangents must be perpendicular at each point

- 1/y' y(1 - x)/(x(x 1))

y' (x(x 1))/(y(x - 1))

separate variables

y dy x(x 1)/(x - 1) dx

y dy (x x)/(x - 1) dx

y dy x 2x/(x - 1) dx

use partial fractions to simplify

2x/(x - 1) 2x/((x 1)(x - 1)) A/(x 1) B/(x - 1) (A B)x (A - B)/((x 1)(x - 1))

A B 0

A - B 2..............A 1, B - 1

x 2x/(x - 1) x 1/(x 1) - 1/(x - 1)

integrate both sides

y/2 x/2 log(x 1) - log(x - 1) D

y x 2 log(x 1)/(x - 1) K

hope it helps

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Find the function f(x) ax^3 bx^2 cx d?

Here the task is to find the value of a,b,

C, and d . And to put these values back in the given expression. f(x)ax^3 bx^2cxd

For x0 ; f(0)d (put x0 on both sides.) Or, -2d (since f(0)-2 given) Ie. d-2

Similarly, for x1 , f(1)abcd Or, 5abc-2 (since f(1)5 given and d-2) Or, abc7...............(1) For x-1, f(-1)-ab-cd (Note!) 3-ab-c-2 ( as d-2 and f(-1)3 ) Or, -ab-c5...........(2)

Adding (1) and (2) : 2b12 Ie. b6

Again for x2: f(2)8a4b2cd Or, 48a4*62c-2 since f(2)4. And d-2 Or, 8a242c6 b6. Obtained above.

Or, 8a2c -18

Or, 4ac-9...................(3)

Adding (2) and (3): 3a b-4 Or, 3a 6-4. (b6) So a-10/3

Put a-10/3 and b6 in (1), then

-10/3 6c7 So, c13/3

Now put the values of a,b, c and d in the given functon f(x)ax^3bx^2cxd f(x)-10/3 x^36x^213/3 x-2.

This is the required form.

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Find a cubic function f(x) ax^3 bx^2 cx d?

Interesting problem. First, a local max or min is where the first derivative is zero (curve is flat).

f'(x) 3ax^2 2bx c, set this to zero and solve using the quadratic equation to yield:

-4, -2 -2b/6a /- sqrt (4b - 12ac)/6a call -2b/6a x and the other part y so you have -4 xy, -2 x-y

so -6 2x or x -3 and x is -2b/6a -3 or 9ab.

Plus this in for b to get 36a^2 81*3a^2 - 12ac or 360a^2 12ac or c 30a.Let's go back to the original two equations and plug these solutions in and solve the two equations for a.

4 a(-4)^3 9a(-4)^2 24a(-4) d

0 a(-2)^3 9a(-2)^2 24a(-2) d (subtract bottom from top)

4 4a or a1, b9, c24. Plug these in to solve for d and you get d20.

So the equation is f(x) x^3 9x^2 24x 20

Hope that helps!

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I'm thinking of trading my 2006 Acura RsX for a Mazda CX-7. Good or bad idea?

Well, in terms of performance... RSX, TSX, Accord In terms of ride comfort.... Accord, TSX, RSX Really depends what you want, i own a RSX and was thinking about getting a TSX.. but its really just a RSX with 4 doors... So i couldnt justify the extra costs for the same feel So then ask yourself.

.. do you want 4 doors or 2.

.. Personally, i would go with the TSX because it doesnt have 2nd gear grind issues like the RSX-S has (all years/models) - TSX has nice things like xeon lights, etc.. I liked the Accord Coupe myself, and if i didnt want 4 doors, it would have been a serious option... But it doesnt handle as nice as your other two options.

.. although it WILL be cheaper to maintain/repair RSX has the smallest rear seats of all your options... the rear seats doesnt fit anyone comfortably... remember this (this is the reason im getting rid of it now) But all your options are pretty good.

.. you wont regret any of them.

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In-Earphones: Shure SE102 vs Sennheiser CX 400/500 II?

In-earphone are a very personal preference. Fit and feel is different for every ear and it also depends on what type of music you are playing through them.

If you have not yet decided one way or another, I would like to toss another brand at you that you in which you might have interest. Ultimate ears makes some great in-earphones. While some of them cost around \$250, one of their models (super.fi 4vi - headset) only costs about \$75 -80. These are some great phones for the price; with punchy bass, smooth mid-range, and crisp highs. They tend to sound a little muffled at ear wrecking volumes, but if you are like me, you like to listen to your music, not be killed by it. For the price, I've never heard a better pair of in-earphones.

Of course, when wanting the best sound, you must have the best recordings. So try to incorporate as many FLAC files in your music collection as possible.

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ax^3 bx^2 cx d. find a,b,c,d given the inflection pt. and relative min.?

We have:

f(x) ax^3 bx^2 cx d

f'(x) 3ax^2 2bx c

f''(x) 6ax 2b

Since there is an inflection point at (0, 0), f''(x) 0 at x 0.

0 6a(0) 2b

> 2b 0

> b 0

Now our function reduces down to:

f(x) ax^3 cx d

f'(x) 3ax^2 c

Since there is a minimum at (2, -4), we have f'(x) 0 at x 2. Then:

0 3a(2)^2 c

> 12a c 0

We know that the function must pass through (0, 0) and (2, -4) as well. So:

f(x) ax^3 cx d

> 0 a(0)^3 c(0) d

> d 0

This now makes our equation:

f(x) ax^3 cx

Plugging in (2, -4):

-4 a(2)^3 c(2)

> 8a 2c -4

> 4a c -2

Since 12a c 0, we have the following system of equations: 12a c 0 4a c -2

We can subtract to get:

8a 2

> a 1/4

Then:

12a c 0

> c -12a

> c -12(1/4)

> c -3

Therefore, a 1/4, b 0, c -3, and d 0.

I hope this helps!

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Show that the system of equations below has a unique solution if and only if bcad: axbys, cxbyt?

If you know about matrix determinants, the determinant of the coefficient matrix is:

| a b |

| c d | ad - bc

...and a linear system needs a nonzero determinant to have a unique solution. ab - cd 0 becomes ab cd in one step.

Another approach is to look at solving the system directly by elimination.

adx bdy ds . . . . multiply 1st equation by d

bcx bdy bt . . . . multiply the 2nd equation by b

(ad - bc)x 0 ds - bt . . . . then subtract to eliminate the y terms

x (ds - bt) / (ad - bc) . . . . and solve for x

That's a unique solution for x when and only when ad - bc 0, therefore ac bd is required for that unique solution to exist.

Now it's your turn. Do the same sort of elimination, starting from the two original equation, but eliminate x this time and solve for y. Do you get an equivalent result, where ad bc is required to make a denominator nonzero?

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Solve ax b cx for x.?

For this, you want the equation to have X on its own on one side so:

x etc

First, you are going to want to get all x's to one side of the equals sign so you can -b from each side, and -cx from each side:

ax b -b -cx cx - cx - b

Simplify:

ax -cx -b

Now, to get x on its own, you need to factorise (ax -cx) which is simple enough because we can divide both by x to get a-c which we write as:

x(a-c) -b

Because x(a-c) x * a x * -c ax -ac

So now that we only have one x, we need to remove (a-c), so to do that we must divide each side by (a-c): (Sorry if all the brackets make it look more confusing)

x(a-c) /(a-c) -b /(a-c)

Simplfyed to:

X -b /(a-c)

Hope that helps! Understanding the answer is more important than the answer itself so feel free to e-mail me if you don't understand anything I explained.

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how do you solve sin(axb)sin(cxd)?

This is kind of silly. You have four degrees of freedom. You can express one parameter, either a, b, c, d, or x, in terms of the others. But there are an uncountable infinitude of answers.

cos() sin(/2 - ),

sin(ax b) sin(/2 - cx - d).

Using the 2-periodicity of the sine function, all you can say is

ax b /2 - cx - d 2n, where n is any integer.

This produces

(ac)x (b d) (4n1)/2.

If a - c anything, then all solutions are of the form

b (4n1)/2 - d, for every integer n and every real number d.

Otherwise, the solutions are

x -(b d)/(a c) (4n1)/(2a2c)

where a, b, c, and d are any real numbers such that a -c and n is any integer.

That's it. Those two cases are all there is (you can substitute "complex number" in for "real number" where it appears.)

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