# Maths differential equation helpWatch

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Thread starter 1 month ago
#1
dy/dx = -(2y/x)
Curve has points, (2,5) and (4,a)
Find a.
Any help would be great. Book says a = 6.25. I keep getting 4.7...
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1 month ago
#2
(Original post by User124)
dy/dx = -(2y/x)
Curve has points, (2,5) and (4,a)
Find a.
Any help would be great. Book says a = 6.25. I keep getting 4.7...
Can you post your working? Must admit, I get 1.25.
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Thread starter 1 month ago
#3
first i collect terms:
dy = -2y(x^-1)dx
(-2y^-1)dy = (x^-1)dx

then i integrate:
-2lny = lnx + c

then i re-arrange:
lny = ln(x^-0.5) + c
y = x^-0.5 + c

then i plug in values to get c. then i use the equation to get a.
but that gets 4.79.
Last edited by User124; 1 month ago
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1 month ago
#4
The 2 should be flipped going from line 1 to 2.
Also, the constant needs to be handled correctly at the end. If the constant was
ln(c)
what could you do to help take inverse logs at the end?

NB if you diff your answer, you do not get the expression in the question. It's close, but ...
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Thread starter 1 month ago
#5
i thought as c is arbitrary it can just be whacked on the end of the equation.
i cant see an issue with:
dy/dx = -2y/x
dy = -2y(x^-1)dx
dy/(-2y) = (x^-1)dx
(-2y^-1)dy = (x^-1)dx
then i integrate...

i am not too sure what is wrong. thanks for replying.

(Original post by mqb2766)
The 2 should be flipped going from line 1 to 2.
Also, the constant needs to be handled correctly at the end. If the constant was
ln(c)
what could you do to help take inverse logs at the end?

NB if you diff your answer, you do not get the expression in the question. It's close, but ...
0
1 month ago
#6
You flip between the 2 on the left hand side being -1/2 or -2. It is -1/2, and so the integral should be sqrt y, not y squared.
Yes the constant is ok, but then you take anti logs of the two additive terms on the right. This isn't ok, combine them into a single term, then remove the logs.
(Original post by User124)
i thought as c is arbitrary it can just be whacked on the end of the equation.
i cant see an issue with:
dy/dx = -2y/x
dy = -2y(x^-1)dx
dy/(-2y) = (x^-1)dx
(-2y^-1)dy = (x^-1)dx
then i integrate...

i am not too sure what is wrong. thanks for replying.
0
Thread starter 1 month ago
#7
please can you write your working. I am struggling to see it. if this is too inconvenient i will ask my teacher. thank you for helping
(Original post by mqb2766)
You flip between the 2 on the left hand side being -1/2 or -2. It is -1/2, and so the integral should be sqrt y, not y squared.
Yes the constant is ok, but then you take anti logs of the two additive terms on the right. This isn't ok, combine them into a single term, then remove the logs.
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1 month ago
#8
(Original post by User124)
please can you write your working. I am struggling to see it. if this is too inconvenient i will ask my teacher. thank you for helping
You said:
first i collect terms:
dy = -2y(x^-1)dx
(-2y^-1)dy = (x^-1)dx this is incorrect as it should be (-1/2) y ^-1 dy = (x^ -1) dx

then i integrate:
-2lny = lnx + c so you need to redo the LHS
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Thread starter 1 month ago
#9
thank you i see it. wow i feel dumb.
(Original post by Muttley79)
You said:
first i collect terms:
dy = -2y(x^-1)dx
(-2y^-1)dy = (x^-1)dx this is incorrect as it should be (-1/2) y ^-1 dy = (x^ -1) dx

then i integrate:
-2lny = lnx + c so you need to redo the LHS
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Thread starter 1 month ago
#10
still dont get a = 6.25 though.
i end up with
y = x^-2 + c
what answers do you get?
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1 month ago
#11
(Original post by User124)
thank you i see it. wow i feel dumb.

Ill try it:

dy/dx = -(2y/x)
dy = -2y(x^-1)dx
(1/-2y) *dy = (x^-1) dx
1/-2 *1/y dy = (x^-1) dx
Integrate gives:
-(1/2)lny = lnx + c
Move over
lny = -2lnx + c
remove ln
y= e^(-2lnx + c)
y= (x^-2)*e^c
y=(x^-2)A where A=e^c

Shove in: 2,5
5=1/4* e^c
20 = e^c
ln20 = c.
(c= 2.9957....)

now shove in 4,a
lny = -2lnx + c
lna = -2ln4 + ln20
This gives a as a number. no matter what you put in.. who wouldve guessed.

Idk man but I know you need to include the e^c bit.. That what I was taught at least.
Last edited by Relentas; 1 month ago
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1 month ago
#12
(Original post by User124)
still dont get a = 6.25 though.
i end up with
y = x^-2 + c
what answers do you get?
Can you post your working? I have kept the 2 on the right as it avoids fractions
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Thread starter 1 month ago
#13
this makes sense. maybe the book is wrong
(Original post by Relentas)
Ill try it:

dy/dx = -(2y/x)
dy = -2y(x^-1)dx
(1/-2y) *dy = (x^-1) dx
1/-2 *1/y dy = (x^-1) dx
Integrate gives:
-(1/2)lny = lnx + c
Move over
lny = -2lnx + c
remove ln
y= e^(-2lnx + c)
y= (x^-2)*e^c
y=(x^-2)A where A=e^c

Shove in: 2,5
5=1/4* e^c
20 = e^c
ln20 = c.
(c= 2.9957....)

now shove in 4,a
lny = -2lnx + c
lna = -2ln4
This gives a as 1.25..

Idk man but I know you need to include the e^c bit.. That what I was taught at least.
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1 month ago
#14
(Original post by Relentas)
Ill try it:

dy/dx = -(2y/x)
dy = -2y(x^-1)dx
(1/-2y) *dy = (x^-1) dx
1/-2 *1/y dy = (x^-1) dx
Integrate gives:
-(1/2)lny = lnx + c
Move over
lny = -2lnx + c
remove ln YOU ARE NOT 'removing' ln - you should substitute in the values then collect ln terms.

Please edit - it's against the rules to post a solution - I would try to collect lns rather than the awkward way you've dealt with THIS LINE,
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1 month ago
#15
(Original post by Muttley79)
Can you post your working? I have kept the 2 on the right as it avoids fractions

Please edit - it's against the rules to post a solution - I would try to collect lns rather than the awkward way you've dealt with THIS LINE,
I saw the first message..

Considering it's apparently not the right answer then I'd personally not consider it to be a solution...
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1 month ago
#16
(Original post by Relentas)
I saw the first message..

Considering it's apparently not the right answer then I'd personally not consider it to be a solution...
Note my comment about the way you've done it - you need to read my comment
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1 month ago
#17
(Original post by Muttley79)
Please edit - it's against the rules to post a solution - I would try to collect lns rather than the awkward way you've dealt with THIS LINE,
Apparently even subbing in the values without any magic gives c= ln50 which still doesn't lead to the right answer. Unless I've done something wrong when calculating, it's 1/2 way there though which I guess is closer
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1 month ago
#18
Fully agree, the answer is 1.25, see post 2.
Edit OP should read mine as well.
(Original post by Muttley79)
Can you post your working? I have kept the 2 on the right as it avoids fractions
Last edited by mqb2766; 1 month ago
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