If you want a bad time, try this P4 integral... Watch

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Aitch
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#1
Report Thread starter 14 years ago
#1
I've just done this. It's from an AL text book, not Heinemann.

If there's a question like this on the paper in June, i shall be fine...

...providing that it is the only question on the paper. It took me forever.

Use the substitution y=ux (u being a function of x) to solve

(x^2)y dy/dx = x^3 + (x^2)y - y^3

confirm

(x^2)y dy/dx = x^3 + (x^2)y - y^3

...it starts off OK...everything cancels nicely... but then...

Aitch
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Aitch
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#2
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Nope.

The answer is of the form

ln((#-#)/(#+#)*(#^#))=2#/(#+#) + c

I had the answer, and it still took me forever...

Clue?

Aitch
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El Stevo
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#3
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i'll work through it... *looks for a pen and paper*
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SsEe
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#4
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Use the substitution y=ux (u being a function of x) to solve

yx²(dy/dx) = x³ + yx² - y³

dy/dx = u + x(du/dx)

ux³(u+x(du/dx)) = x³ + ux³ - u³x³
u² + ux(du/dx) = 1 + u - u³
ux(du/dx) = 1 + u - u² - u³
Separating out the variables we get:
-int[u/(u³+u²-u-1)]du = int(1/x)dx
-int[u/[(u-1)(u+1)²]]du = int(1/x)dx
Using partial fractions we get:
int[1/(u-1) - 1/(u+1) + 2/(u+1)²]du = -4 int(1/x)dx
ln(u-1) - ln(u+1) - 2/(u+1) = -4ln(x) - lnA = (where -lnA is the arbitary constant)
After some rearranging you get:

(Ayx^4 - Ax^5)/(y+x) = e^(2x/(y+x))

================================ ===================
ln((#-#)/(#+#)*(#^#))=2#/(#+#) + c

ln(u-1) - ln(u+1) - 2/(u+1) = -4ln(x) + c

ln((u-1)/(u+1)*(x^4))=2/(u+1) + c
ln((y-x)/(y+x)*(x^4))=2x/(x+y) + c
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El Stevo
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#5
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gah. beat me to posting it... i took way too long... 45 minutes... we don't have those in p4, i don't think...
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Aitch
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#6
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(Original post by SsEe)
Use the substitution y=ux (u being a function of x) to solve

yx²(dy/dx) = x³ + yx² - y³

dy/dx = u + x(du/dx)

ux³(u+x(du/dx)) = x³ + ux³ - u³x³
u² + ux(du/dx) = 1 + u - u³
ux(du/dx) = 1 + u - u² - u³
Separating out the variables we get:
-int[u/(u³+u²-u-1)]du = int(1/x)dx
-int[u/[(u-1)(u+1)²]]du = int(1/x)dx
Using partial fractions we get:
int[1/(u-1) - 1/(u+1) + 2/(u+1)²]du = -4 int(1/x)dx
ln(u-1) - ln(u+1) - 2/(u+1) = -4ln(x) - lnA = (where -lnA is the arbitary constant)
After some rearranging you get:

(Ayx^4 - Ax^5)/(y+x) = e^(2x/(y+x))

================================ ===================
ln((#-#)/(#+#)*(#^#))=2#/(#+#) + c

ln(u-1) - ln(u+1) - 2/(u+1) = -4ln(x) + c

ln((u-1)/(u+1)*(x^4))=2/(u+1) + c
ln((y-x)/(y+x)*(x^4))=2x/(x+y) + c
Yup - same method as I used.

It was sorting out the partial fractions that slowed me down.

Aitch
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Gauss
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#7
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p4? jeez, der was nothing like dat when i did p4. it looks like some m1m1 stuff.

euclid
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Aitch
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#8
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(Original post by Euclid)
p4? jeez, der was nothing like dat when i did p4. it looks like some m1m1 stuff.

euclid
The question is from Further Pure Maths by Gaulter & Gaulter, so deals with P4-6 topics. I needed more practice with FOD and SOD equations, especially when sustitutions were involved, so just chose some qs from the appropriate chapter. Could be P5-6? I don't know the content of these.

Aitch
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Aitch
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#9
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Explanation required?

When future generations of mathematicians are tired of referring to first order differential equations and second order differential equations, fods and sods will become acceptable terminology. You mark my worms.

Aitch
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Nick Sutcliffe
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#10
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#10
That's P4, I've not done any P5 or 6 yet, but could do that problem (although made a stupid mistake on 4th line of working, missed off a "u" on multiplication making things rather more difficult . Or to be fair, incorrect). Point being, I knew what I was doing with P4 knowledge.
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RObTRIP
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#11
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yup thats la creme de la creme of p4 second order differential equations
i quite enjoy them..

well compared to m4/m5!! :bootyshak
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El Stevo
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#12
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(Original post by RObTRIP)
yup thats la creme de la creme of p4 second order differential equations
i quite enjoy them..

well compared to m4/m5!! :bootyshak
okay, let me clarify. i could do it, but it isn't in AQA Maths A p4...
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