Separable differential equation

ch2kb0x
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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.
 
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Start with u = y/x, or equivalently, y = ux. From this, find dy/dx.
 
Okay, so since you said y = ux, I am thinking that this is a homogenous equation...

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 I am wrong.
 
ch2kb0x said:
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.

Well, that is because it isn't a separateable equation... It isn't even an ODE. Sure it's right?
 
Yeah, copied exactly from textbook.
 
ch2kb0x said:
Okay, so since you said y = ux, I am thinking that this is a homogenous equation...

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 I am wrong.
You can write the equation as dy/dx = 1/(x + 2y), where the right side is a function of x and y. I'm just offering a suggested approach based on your first post. It may or may not work.
 
Yeah, it is a function of x and y, but I don't think it's in the g(y/x) form.
 
Mark44 said:
You can write the equation as dy/dx = 1/(x + 2y), where the right side is a function of x and y. I'm just offering a suggested approach based on your first post. It may or may not work.

To me it seems that you are suggesting a linear trial solution?
 
The OP said he wanted to put this into g(y/x) form, so that made me think of the substitution I suggested. I don't have access to my DE textbooks right now, so I don't have any more ideas on solving this one.
 
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