- #1

- 4

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Solve showing full working

(x^2 -1)y' + 2xy = x

You must show full working.

(x^2 -1)y' + 2xy = x

You must show full working.

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- Thread starter josek6
- Start date

- #1

- 4

- 0

Solve showing full working

(x^2 -1)y' + 2xy = x

You must show full working.

(x^2 -1)y' + 2xy = x

You must show full working.

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- #2

Homework Helper

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Try to rearrange and find an integrating factor u such that

u(x^2 -1)y' + u(2xy - x)

is and exact differential

- #3

- 4

- 0

So I have to rewrite the equation in the standard form y' + Py =Q

IF = e ^ ∫P.dx

where P = 2x ?

- #4

- 4

- 0

Thanks...I solved it my solution is

y(x^2 -1)=(x^2)/2

y(x^2 -1)=(x^2)/2

- #5

Homework Helper

- 2,461

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Good, but you need the arbitrary constant

y(x^2 -1)=(x^2)/2+C

y(x^2 -1)=(x^2)/2+C

- #6

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- #7

- 2

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(x^2-1)y = \int{x};-->

(x^2-1)y = 0.5 x^2 + C

y = (0.5 x^2 + C) / ( x^2 - 1 )

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