Trouble separating equation (differential equation)

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Homework Help Overview

The problem involves a differential equation of the form y' = (2x) / (y + (x^2)y) with the initial condition y(0) = -2. The original poster expresses difficulty in separating the variables x and y.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss attempts to manipulate the equation to separate variables, with one participant suggesting a specific rearrangement. There is also a focus on the algebraic manipulation involved in the separation process.

Discussion Status

Some participants have offered guidance on the algebraic steps needed to separate the variables, while others have pointed out potential mistakes in the manipulation of terms. The discussion reflects a mix of interpretations and attempts to clarify the algebra involved.

Contextual Notes

There are mentions of initial conditions and concerns about algebraic reasoning in the context of solving differential equations. The original poster's approach and reasoning are questioned, particularly regarding the handling of constants and logarithmic terms.

darryw
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Homework Statement


y' = (2x) / y+(x^2)y y(0) = -2
i realize this is sort of algebra prob, but i can't seem to separate x's from y's..



Homework Equations





The Attempt at a Solution

 
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darryw said:

Homework Statement


y' = (2x) / y+(x^2)y y(0) = -2
i realize this is sort of algebra prob, but i can't seem to separate x's from y's..



Homework Equations





The Attempt at a Solution


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

Now try separating it.
 
wow. that shouldve been obvious.
thanks.
 
Please check my work thanks a lot..
find solution in explicit form:

y' = (2x) / y+(x^2)y initial conditions: y(0) = -2

dy/dx = 2x/y(1+x^2)

y dy = 2x + (2/x) dx

integrate both sides...

(1/2)y^2 = x^2 + 2ln|x| + c

y^2 = 2x^2 + 4ln|x| + 2c

y = root [2x^2 + 4ln|x| + c]

apply IC:

-2 = root [2(0)^2 + 4ln|0| + c]

-2 = (root 4) + c

c = -4 (i am reasong that root c is same as c because whether its root c or just c, both are still constants.. right?)

so my solution is:
y = root [2x^2 + 4ln|x| -4]
 
darryw said:
Please check my work thanks a lot..
find solution in explicit form:

y' = (2x) / y+(x^2)y initial conditions: y(0) = -2

dy/dx = 2x/y(1+x^2)

y dy = 2x + (2/x) dx
Mistake in line above. It looks like you are trying to say that 2x/(1 + x^2) = 2x/1 + 2x/x^2.

That's just like saying that 2/(4 + 8) = 2/4 + 2/8 = 1/2 + 1/4 = 3/4, which I hope you can see is not true.

If you are working with differential equations, you shouldn't be having problems with basic algebra manipulation.
darryw said:
integrate both sides...

(1/2)y^2 = x^2 + 2ln|x| + c

y^2 = 2x^2 + 4ln|x| + 2c

y = root [2x^2 + 4ln|x| + c]

apply IC:

-2 = root [2(0)^2 + 4ln|0| + c]

-2 = (root 4) + c

c = -4 (i am reasong that root c is same as c because whether its root c or just c, both are still constants.. right?)

so my solution is:
y = root [2x^2 + 4ln|x| -4]
 

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