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Reply 20
Original post by SubAtomic



4. dydx=(1y2)x1x2\dfrac{dy}{dx} = \dfrac{(1-y^2)x}{\sqrt{1-x^2}}

Hmmm had a go but think I may be going wrong somewhere.

So I multiplied by the denominator, did reciprocal, multiplied by x and that is where stopped as am not sure if am going wrong

(1x2)1/2dydx=(1y2)x\displaystyle (1 – x^2)^{1/2} \dfrac{dy}{dx} = (1 – y^2)x

x1x2dx=11y2dy\displaystyle \dfrac{x}{\sqrt{1–x^2}} dx = \dfrac{1}{\sqrt{1–y^2}} dy


dydx=(1y2)x1x2    11y2 dy=x1x2 dx    1(1y)(1+y) dy=x1x2 dx\displaystyle \dfrac{dy}{dx} = \dfrac{(1-y^2)x}{\sqrt{1-x^2}} \implies \int \frac1{1-y^2} \ dy = \int \frac{x}{\sqrt{1-x^2}} \ dx \\ \implies \int \frac1{(1-y)(1+y)} \ dy = \int \frac{x}{\sqrt{1-x^2}} \ dx

Use partial fractions for LHS.

For RHS, x1x2 dx    x(1x2)12 dx \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx \implies \int x(1-x^2)^{-\frac12} \ dx

To integrate this, try differentiating, z=(1x2)12 \displaystyle z=(1-x^2)^{\frac12}
Reply 21
Original post by raheem94
=

To integrate this, try differentiating, z=(1x2)12 \displaystyle z=(1-x^2)^{\frac12}


I would let z= 1-x^2.

It's even easier to manipulate it so that it's of the form: f'(x)/sqrtf(x)=

2sqrtf(x)+c.
Reply 22
Original post by SubAtomic


Edit: f1mad -2x


That's right, and you already have an x present in the integrand.

So?
Reply 23
Original post by SubAtomic
Lol no idea why I put the root sign in the (1-y) expression, had a mini latex meltdown

ddx(1x2)1/2=x1x2\displaystyle\dfrac {d}{dx} (1-x^2)^{1/2} = - \dfrac{x}{\sqrt{1-x^2}}



Divide both sides by -1, and integrate both sides.
Reply 24
Original post by SubAtomic
Lol no idea why I put the root sign in the (1-y) expression, had a mini latex meltdown

ddx(1x2)1/2=x1x2\displaystyle\dfrac {d}{dx} (1-x^2)^{1/2} = - \dfrac{x}{\sqrt{1-x^2}}


This shows us, x1x2 dx=(1x2)12+C \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx = -(1-x^2)^{\frac12} + C
(edited 11 years ago)
Reply 25
Original post by raheem94
This shows us, x1x2 dx=(1x2)+C \displaystyle \int \frac{x}{\sqrt{1-x^2}} \ dx = -(1-x^2) + C


It's not.
Reply 26
Original post by SubAtomic
Put -1/2 outside integrand and -2x as numerator?

Lost:s-smilie:


Sorry, i made a mistake, forgot to type the power, i have corrected it now.
Reply 27
Original post by f1mad
It's not.


Thanks for correcting me, i was replying to multiple threads, along with making some grade boundary calculations, hence i was doing everything very quickly. I did it correctly myself but while typing i forgot to write the power.
Reply 28
Original post by SubAtomic
Will go work on what have got and come back later as feel a bit lost:rolleyes:


I x/(1-x^2)^1/2 dx

Perhaps following on with the substitution would better sense.

Let u= 1-x^2

du/dx= -2x

du= -2x dx

We already have x in the integrand, so divide by -2

du/-2= xdx

Then it becomes: I -1/2 * 1/(u^1/2) du

You will see that this is a standard integral result:

I f'(x)/sqrtf(x) dx=

2sqrtf(x)+c.
(edited 11 years ago)
Reply 29
Original post by f1mad
I would let z= 1-x^2.

It's even easier to manipulate it so that it's of the form: f'(x)/sqrtf(x)=

2sqrtf(x)+c.


At first i solved this by using a trigonometric substitution, x=sinθ x=sin\theta .

Though it will be difficult for the OP to understand this.
Reply 30
So am back and as I suspected I have not done anything partial fraction. I have been using the Integral formulas.

So

x(1x2)1/2 dx=12(1x2)1/2(2x) dx\displaystyle\int x(1-x^2)^{-1/2} \ dx = - \dfrac{1}{2} \int (1-x^2)^{-1/2}(-2x) \ dx

So this becomes

12×21(1x2)1/2+C\displaystyle - \dfrac{1}{2} \times \dfrac{2}{1} (1-x^2)^{1/2} + C

So will start again

(1y2)1 dy=\displaystyle\int (1-y^2)^{-1} \ dy =

Have never done partial fractions is this the only way? If so I need to go learn the method.
(edited 11 years ago)
Reply 31
Original post by SubAtomic
So am back and as I suspected I have not done anything partial fraction. I have been using the Integral formulas.

So

x(1x2)1/2 dx=12(1x2)1/2(2x) dx\displaystyle\int x(1-x^2)^{-1/2} \ dx = - \dfrac{1}{2} \int (1-x^2)^{-1/2}(-2x) \ dx

So this becomes

12×21(1x2)1/2+C\displaystyle - \dfrac{1}{2} \times \dfrac{2}{1} (1-x^2)^{1/2} + C

And for the y term the same scenario

12(1y2)2(2y) dy\displaystyle - \dfrac{1}{2} \int (1-y^2)^{-2}(-2y) \ dy

And then

12×1(1y2)1 = 121(1y2)\displaystyle - \dfrac{1}{2} \times -1 (1-y^2)^{-1} \ = \ \dfrac{1}{2} \cdot \dfrac{1}{(1-y^2)}

So

12(1y2) = (1x2)1/2+C\displaystyle\dfrac{1}{2(1-y^2)} \ = \ -(1-x^2)^{1/2} + C

Will edit this in a sec back to the paper for a min


Your 'x' terms are correct, but 'y' is wrong.

Here is how to deal with the 'y' part.

11y2 dy=1(1y)(1+y) dy \displaystyle \int \frac1{1-y^2} \ dy = \int \frac1{(1-y)(1+y)} \ dy

We can't integrate this unless we use partial fractions.

1(1y)(1+y)A1y+B1+y1(1y)(1+y)A(1+y)+B(1y)(1y)(1+y) \displaystyle \frac1{(1-y)(1+y)} \equiv \frac{A}{1-y} + \frac{B}{1+y} \\ \displaystyle \frac1{(1-y)(1+y)} \equiv \frac{A(1+y)+B(1-y)}{(1-y)(1+y)}

So solving the above expression gives, A=12 and B=12 A=\frac12 \text{ and } B=\frac12

Hence, 1(1y)(1+y)121y+121+y \displaystyle \frac1{(1-y)(1+y)} \equiv \frac{\frac12}{1-y} + \frac{\frac12}{1+y}

So the integral becomes,
11y2 dy=1(1y)(1+y) dy=(121y+121+y) dy=12(11y+11+y) dy=12(ln(1y)+ln(1+y)) \displaystyle \int \frac1{1-y^2} \ dy = \int \frac1{(1-y)(1+y)} \ dy = \int \left( \frac{\frac12}{1-y} + \frac{\frac12}{1+y} \right) \ dy =\frac12 \int \left( \frac{1}{1-y} + \frac{1}{1+y} \right) \ dy = \frac12(-ln(1-y)+ln(1+y))

Do you get it now?
Reply 32
Original post by SubAtomic
Kinda. Are partial fractions the only way because if this is the case then I need to go learn the method.


Yes, you need to know partial fractions for it. There might be some other way, but i can't think of any other way.

Do you really don't know partial fractions?

You should learn it before doing C4 integration.
Reply 33
Original post by SubAtomic
Am not doing a standard A-level it is a supposed equivalent (although it seems to be leaving me ****ed on questions like this). Need to go learn it then.

Examsolutions?


Which qualification are you doing?

Examsolutions is a good website to learn A-Level maths, here is a link to partial fractions, http://examsolutions.co.uk/maths-revision/core-maths/algebra-and-functions/rational-expressions/partial-fractions/intoduction.php
Reply 34
Original post by SubAtomic
Well I'm supposedly done lol :rolleyes: I have near enough completed the course, just on with the hypothesis tests and that's it.

Here have a look at a taster for yourself here and then tell me what you think. I don't think it quite cuts it as an A-level equivalent.

Cheers for the link


I may have a look at the link later, i am currently busy studying Hypothesis testing, it is a bit difficult to understand at first.
Reply 35
Quick question.

So a differential equation I have to find

dydx=(ysomethingsomethingx)somethingsomething\displaystyle \frac {dy}{dx} = (y^{something} \cdot something^{x})^\frac{something}{something}

And am given y=50 when x=0 so y(0)=50 y = 50 \ when \ x = 0 \ so \ y(0) = 50

Do I integrate and simplify as much as possible before subbing x = 0 or not? Bit confused with an assignment question still because I get the same answer over and over again and cannot seem to manipulate my answer to look like the choices I have.

lol sorry for all the somethings:rolleyes:
(edited 11 years ago)
Reply 36
Original post by SubAtomic
Quick question.

So a differential equation I have to find

dydx=(ysomethingsomethingx)somethingsomething\displaystyle \frac {dy}{dx} = (y^{something} \cdot something^{x})^\frac{something}{something}

And am given y=(something) when x=0 so y(0)=something y = (something) \ when \ x = 0 \ so \ y(0) = something

Do I integrate and simplify as much as possible before using x = 0 or not? Bit confused with an assignment question still because I get the same answer over and over again and cannot seem to manipulate my answer to look like the choices I have.

lol sorry for all the somethings:rolleyes:


Can you post the exact question.

I can't assume what 'something' is.
Reply 37
Original post by SubAtomic
...


You can only plug in what x and y are, when you have got the GS of the DE.

You don't really need to simplify it once that's done, unless they ask for it. Although usually you would get it into the form y(x)= something.

Once you have got the GS, plug in x=0, y=50 and solve for the arbitrary constant to get a particular solution.
(edited 11 years ago)
Reply 38
Original post by f1mad
You can only plug in what x and y are, when you have got the GS of the DE.

You don't really need to simplify it once that's done, unless they ask for it. Although usually you would get it into the form y(x)= something.

Once you have got the GS, plug in x=0, y=50 and solve for the arbitrary constant to get a particular solution.


Cheers get ya:cool:

Original post by f1mad
...



So I would use x = 0 in the general solution then simplify to y = something after? Because I definitely have to get it as y = something in the end, although this is what have been doing so I need to manipulate and manipulate till I get the answer then:s-smilie:
(edited 11 years ago)
Reply 39
Original post by SubAtomic
Cheers get ya:cool:




So I would use x = 0 in the general solution then simplify to y = something after? Because I definitely have to get it as y = something in the end, although this is what have been doing so I need to manipulate and manipulate till I get the answer then:s-smilie:


You could do either.

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