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# P4 Second Order Differential Equations Question watch

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1. Find the general solution of

x d2y - 2 dy + x = 0
dx2 dx

Using the substitution dy/dx = v

(Heinemann P4 textbook, Ex6E, Q7)

Any takers?
2. im gonna try it now... might take a little while though... so if someone hasn't posted their answer i will post mine soon
3. I tried and failed - sorry!
4. ok i think i got it... but i left the book downstairs... is the answer:
y=(x^2)/2 + Ax^3 +B

forgot to check before
5. (Original post by S1M)
ok i think i got it... but i left the book downstairs... is the answer:
y=(x^2)/2 + Ax^3 +B

forgot to check before
Hooray! Now could you please enlighten me as to how you did it, because I don't have a clue!
6. ok this might take some time...

dy/dx=v
d2y/dx2 = dv/dx
substitute that into eqn

xdv/dx - 2v + x = 0

divide by c

dv/dx -2v/x +x/x = 0

dv/dx + (- 2/x)v = -1

IF = e^(int of -2/x) = e ^ (-2lnx) = e ^ (ln(x^-2))

then from there its just like 1st order
7. ok ill finish it off for ya...

v/x^2 = int 1/x^2
= 1/x + C
then sub back v = dy/dx

rearrange
dy/dx = x + Cx^2
intergrate:
y = (x^2)/2 + (C/3)x^3 + B
C/3 = A
and you must be able to do the last step
8. (Original post by S1M)
ok this might take some time...

dv/dx -2v/x +x/x = 0

dv/dx + (- 2/x)v = 1
Shouldn't that be -1?
9. geehhh I'm stupid - I forgot the +c when I integrated the -1/x^2. Talk about dumb...

Cheers S1M, rep cometh your way!
10. (Original post by shift3)
Shouldn't that be -1?
yup -1... was just a typo when copying it from paper...
11. (Original post by flyinghorse)
geehhh I'm stupid - I forgot the +c when I integrated the -1/x^2. Talk about dumb...

Cheers S1M, rep cometh your way!
Your welcome... thanks for the rep
12. well i was nearly there

i just forgot to integrate to get y = ... thanks anyway thats good work!
13. (Original post by kikzen)
well i was nearly there

i just forgot to integrate to get y = ... thanks anyway thats good work!
no problem... thanks for the rep

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