- #1

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## Homework Statement

(x + 2y) dy/dx = 1, y(0) = 1

## Homework Equations

## The Attempt at a Solution

Problem is, I can't separate it. This might be a homogenous type? If so, how would I make it into the g(y/x) form.

Thank you.

- Thread starter ch2kb0x
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- #1

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(x + 2y) dy/dx = 1, y(0) = 1

Problem is, I can't separate it. This might be a homogenous type? If so, how would I make it into the g(y/x) form.

Thank you.

- #2

Mark44

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Start with u = y/x, or equivalently, y = ux. From this, find dy/dx.

- #3

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However, if it is a homogeneous equation, before we can plug in y = ux, aren't we suppose to first have the equation in the form of dy/dx = f(x,y), where there exists a function such that f(x,y) is expressed g(y/x).

Then, AFTEr we can do the y=ux thing. correct me if Im wrong.

- #4

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Well, that is because it isn't a separateable equation... It isn't even an ODE. Sure it's right?Problem is, I can't separate it. This might be a homogenous type? If so, how would I make it into the g(y/x) form.

Thank you.

- #5

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Yeah, copied exactly from textbook.

- #6

Mark44

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You can write the equation as dy/dx = 1/(x + 2y), where the right side

However, if it is a homogeneous equation, before we can plug in y = ux, aren't we suppose to first have the equation in the form of dy/dx = f(x,y), where there exists a function such that f(x,y) is expressed g(y/x).

Then, AFTEr we can do the y=ux thing. correct me if Im wrong.

- #7

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Yeah, it is a function of x and y, but I don't think it's in the g(y/x) form.

- #8

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To me it seems that you are suggesting a linear trial solution?You can write the equation as dy/dx = 1/(x + 2y), where the right sideisa function of x and y. I'm just offering a suggested approach based on your first post. It may or may not work.

- #9

Mark44

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