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# Differential Equations Question watch

1. Hi!
I have become a bit stuck on this question the C3C4 Elmwood textbook.

Given x and y are positive, dy/dx +y/x^2 =0 and y= e when x=1, show that the solution may be written in the form y=e^(x-1)

My attempt:
dy/dx = -y/x^2
integral of 1/y with respect to y = integral of -1/x^2 with respect to x
lny = 1/3x^(-3) +c
y= Ae^(1/3x^(-3))
When y=e and x=1
e = Ae^(1/3)
A=e / e^(1/3)
A=e^(1-1/3)
A=e^(2/3)

Therefore
y=e^(1/3)*e^(1/3x^(-3))
y=e^(2/9x^3)
Which is not what I am supposed to get :/

Any suggestions to where I have gone wrong?
2. (Original post by Ruineth)
Hi!
I have become a bit stuck on this question the C3C4 Elmwood textbook.

Given x and y are positive, dy/dx +y/x^2 =0 and y= e when x=1, show that the solution may be written in the form y=e^(x-1)

My attempt:
dy/dx = -y/x^2
integral of 1/y with respect to y = integral of -1/x^2 with respect to x
lny = 1/3x^(-3) +c
y= Ae^(1/3x^(-3))
When y=e and x=1
e = Ae^(1/3)
A=e / e^(1/3)
A=e^(1-1/3)
A=e^(2/3)

Therefore
y=e^(1/3)*e^(1/3x^(-3))
y=e^(2/9x^3)
Which is not what I am supposed to get :/

Any suggestions to where I have gone wrong?
please post a photo of the workings
3. (Original post by TeeEm)
please post a photo of the workings
Of course
4. (Original post by Ruineth)
Of course
I am off to work now but someone will help you real soon

All the best !!
5. (Original post by Ruineth)
Of course
When integrating you need to raise the power by one and then divide, careful with the negative power - see your third and fourth lines of working.
6. (Original post by GingerGnu)
When integrating you need to raise the power by one and then divide, careful with the negative power - see your third and fourth lines of working.
Ah yes! Got it, stupid mistakes...

Thank you so much

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