- #1

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[tex]x \frac{dy}{dx}+x^2y+y=0[/tex]

to this

[tex]\frac{dy}{dx}+(x+\frac{1}{x})y=0[/tex]

this is what I would of thought it would be...

[tex]\frac{dy}{dx}+(x+\frac{y}{x})+\frac{y}{x}=0[/tex]

no?

- Thread starter snowJT
- Start date

- #1

- 118

- 0

[tex]x \frac{dy}{dx}+x^2y+y=0[/tex]

to this

[tex]\frac{dy}{dx}+(x+\frac{1}{x})y=0[/tex]

this is what I would of thought it would be...

[tex]\frac{dy}{dx}+(x+\frac{y}{x})+\frac{y}{x}=0[/tex]

no?

- #2

cristo

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Divide the original equation by x; what do you get?

- #3

arildno

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Factorize [itex]x^{2}y+y[/itex] in the simplest manner.

What do you get?

What do you get?

- #4

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you get [tex]\frac{dy}{dx}+(\frac{x^2}{x}+\frac{y}{x})+\frac{y}{x}=0[/tex]

Divide the original equation by x; what do you get?

you get [tex]\frac{x^2}{y}+1[/tex]Factorize [itex]x^{2}y+y[/itex] in the simplest manner.

What do you get?

- #5

arildno

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No, you don't.

What does it mean to FACTORIZE?

What does it mean to FACTORIZE?

- #6

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oh I see now... thanks

but why are you not dividing the y thats on the most right side of the LHS by x?

- #7

arildno

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[tex]x\frac{dy}{dx}+(x^{2}+1)y=0[/tex]

Multiply this equation with 1/x; what do you get?

- #8

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sorry, I see it now... I'm just really bad at these things...

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