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FP2: general solution of an equation - Where did I go wrong?

Should've gotten 2+-1/x root(3x^2 + A)

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Original post by creativebuzz
Should've gotten 2+-1/x root(3x^2 + A)



You didn't integrate the fraction on the LHS correctly.

Differentiate

−2ln⁡(u2−4u+1)\displaystyle -2\ln(u^2 - 4u +1) to see what I mean :smile:
Original post by Indeterminate
You didn't integrate the fraction on the LHS correctly.

Differentiate

−2ln⁡(u2−4u+1)\displaystyle -2\ln(u^2 - 4u +1) to see what I mean :smile:


Oh, I thought it was right because differentiation u^2 - 4u + 1 is the same as 2- u if you multiply the top by -2
Original post by creativebuzz
Oh, I thought it was right because differentiation u^2 - 4u + 1 is the same as 2- u if you multiply the top by -2


If you want to multiply the top of the fraction by -2 (to make it look like the differential of the bottom) then you'll have to undo it after integrating. :smile:
Original post by creativebuzz
Oh, I thought it was right because differentiation u^2 - 4u + 1 is the same as 2- u if you multiply the top by -2
I can't think of a way of making the error clear without being patronizing, but what you've said is analogous to:

"I thought 3/6 = 2 because 3 is the same as 6 if you multiply the top by 2"

(and I think it's worth pointing out because you do need to (correctly) do this kind of mental adjustment between "what you want the numerator to be" and "what it is").
Original post by Indeterminate
If you want to multiply the top of the fraction by -2 (to make it look like the differential of the bottom) then you'll have to undo it after integrating. :smile:


Oh dear, I can't believe I forgot something like that :P That's the issue with doing c4 in your first year and fp2 in your 2nd.. you've forgotten the rules required for integration!

While I try to give this question another shot, would you mind giving me a hand on this super quick question? I was going to solve it using the standard fp2 inequalities method but I don't know how that answers the question..

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Original post by creativebuzz
Oh dear, I can't believe I forgot something like that :P That's the issue with doing c4 in your first year and fp2 in your 2nd.. you've forgotten the rules required for integration!

While I try to give this question another shot, would you mind giving me a hand on this super quick question? I was going to solve it using the standard fp2 inequalities method but I don't know how that answers the question..



Multiply through by

∣x2+2x+2∣|x^2 + 2x + 2|

(which you're allowed to do because x2+2x+2=(x+1)2+1>0x^2 + 2x + 2 = (x+1)^2 + 1 > 0 for all x)

Next I'd recommend sketching the graphs of each side of the inequality. That way you'll be able to see the set of values that you're after.
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Gome44
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Original post by creativebuzz
Oh dear, I can't believe I forgot something like that :P That's the issue with doing c4 in your first year and fp2 in your 2nd.. you've forgotten the rules required for integration!

While I try to give this question another shot, would you mind giving me a hand on this super quick question? I was going to solve it using the standard fp2 inequalities method but I don't know how that answers the question..



Multiply through by |x^2 +2x +2| and sketch the graphs. 0.5|x^2 +2x+2| should be very easy to sketch (I'll let you figure out why) :smile:
Original post by Indeterminate
Multiply through by

∣x2+2x+2∣|x^2 + 2x + 2|

(which you're allowed to do because x2+2x+2=(x+1)2+1>0x^2 + 2x + 2 = (x+1)^2 + 1 > 0 for all x)

Next I'd recommend sketching the graphs of each side of the inequality. That way you'll be able to see the set of values that you're after.


So you're saying that I should do the "generic fp2 inequality method"?

I gave the other question another shot and I still can't seem to get the correct answer :/

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Original post by creativebuzz
So you're saying that I should do the "generic fp2 inequality method"?

I gave the other question another shot and I still can't seem to get the correct answer :/



You're very close, but you've just made 2 minor mistakes. Firstly, you should've had

u2−4u+1=1A2x2u^2 - 4u +1 = \dfrac{1}{A^2x^2}

a few lines before the end (since you squared both sides).

Then all you had to do was set

1A2=B\dfrac{1}{A^2} = B

which would have given you

u=2±3+Bx2\displaystyle u = 2 \pm \sqrt{3 + \dfrac{B}{x^2}}

Now, notice that you can't just multiply the inside of the square root by x^2 because that will change what you're taking the square root of. You need to multiply the inside by

x2x2\dfrac{x^2}{x^2}

instead :smile:
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TeeEm
19
Original post by creativebuzz
So you're saying that I should do the "generic fp2 inequality method"?

I gave the other question another shot and I still can't seem to get the correct answer :/



it look like somebody else has found your errors
Original post by Indeterminate
You're very close, but you've just made 2 minor mistakes. Firstly, you should've had

u2−4u+1=1A2x2u^2 - 4u +1 = \dfrac{1}{A^2x^2}

a few lines before the end (since you squared both sides).

Then all you had to do was set

1A2=B\dfrac{1}{A^2} = B

which would have given you

u=2±3+Bx2\displaystyle u = 2 \pm \sqrt{3 + \dfrac{B}{x^2}}

Now, notice that you can't just multiply the inside of the square root by x^2
because that will change what you're taking the square root of. You need to multiply the inside by

x2x2\dfrac{x^2}{x^2}

instead :smile:

But if you multiplied the inside of the square root by x^2/X^2

wouldn't you just get



because 3x^2/x^2 = 3 and Bx^2/x^4 = B/x^2
Original post by creativebuzz
But if you multiplied the inside of the square root by x^2/X^2

wouldn't you just get



because 3x^2/x^2 = 3 and Bx^2/x^4 = B/x^2


No

u=2±1x2(3x2+B)\displaystyle u = 2 \pm \sqrt{\dfrac{1}{x^2}\left(3x^2 + B\right)}

Can you see it now? :smile:
Original post by Indeterminate
No

u=2±1x2(3x2+B)\displaystyle u = 2 \pm \sqrt{\dfrac{1}{x^2}\left(3x^2 + B\right)}

Can you see it now? :smile:


Oh I see how that will follow onto my answer!

But isn't that factoring out 1/x^2 as opposed to multiplying top and bottom by x^2?
Original post by creativebuzz
Oh I see how that will follow onto my answer!

But isn't that factoring out 1/x^2 as opposed to multiplying top and bottom by x^2?


Well, I suppose could call it either of those things. This is because, by introducing 1x2\frac{1}{x^2}, you've compensated for multiplying by x2x^2. Those operations combine to make x2x2\dfrac{x^2}{x^2}.

Of course, everyone will have a different way of looking at it :smile:
Original post by Indeterminate
Well, I suppose could call it either of those things. This is because, by introducing 1x2\frac{1}{x^2}, you've compensated for multiplying by x2x^2. Those operations combine to make x2x2\dfrac{x^2}{x^2}.

Of course, everyone will have a different way of looking at it :smile:


Ah okay :smile:

The next part of the question is to deduce that large values of x, f(x) can be approximated by the expression (2 + root3)x

the only way I can see them getting there is by ignoring all the constants as they're not "values of x" and then after squaring and factorising you'd get (2 + root3)x but that doesn't seem like a "legit method" to go by doing it?
Original post by creativebuzz
Ah okay :smile:

The next part of the question is to deduce that large values of x, f(x) can be approximated by the expression (2 + root3)x

the only way I can see them getting there is by ignoring all the constants as they're not "values of x" and then after squaring and factorising you'd get (2 + root3)x but that doesn't seem like a "legit method" to go by doing it?


Considering that this is just A-level stuff, all you'd be expected to say is stuff like

For large values of x, 3x2+B3x^2 + B can be approximated by 3x23x^2

And go from there :smile:


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(edited 8 years ago)
Original post by creativebuzz
Oh dear, I can't believe I forgot something like that :P That's the issue with doing c4 in your first year and fp2 in your 2nd.. you've forgotten the rules required for integration!

While I try to give this question another shot, would you mind giving me a hand on this super quick question? I was going to solve it using the standard fp2 inequalities method but I don't know how that answers the question..



(|x + 1| - 1)^2 >= 0, because squares are always positive.
So |x + 1|^2 -2|x + 1| + 1 >= 0,
or |x+1|^2 +1 >= 2|x+1|.
but |x+1|^2 +1 = |(x+1)^2 +1| = |x^2+2x+2|, so |x^2+2x+2|>=2|x+1|, and the desired inequality follows.
Original post by Indeterminate
Considering that this is just A-level stuff, all you'd just be expected to say is stuff like

For large values of x, 3x2+B3x^2 + B can be approximated by 3x23x^2

And go from there :smile:


Posted from TSR Mobile


Got it, thank you! :smile:

Could you spot where I went wrong in this "find the particular solution" question?



That question was part b to question 13, so this is 13a if you need to look at it (just in case my error was made there):


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Zacken
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Original post by creativebuzz

x


Haven't checked it out, but I'm puzzled over why the argument of the exponential function suddenly changed from -3x to -4x in the fourth line.

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