Solve this differential equation using separation of variables

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Homework Help Overview

The discussion revolves around solving a first-order ordinary differential equation using separation of variables. The equation presented is dy/dx = y^2 x, with an initial condition of y = -1 when x = 0.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the separation of variables technique, with some attempting to integrate the separated equation. Questions arise regarding the transformation of y^2 into 1/y^2 during the separation process, as well as the interpretation of coefficients in the equation.

Discussion Status

The discussion is active, with participants providing guidance on the separation of variables and integration steps. There is a focus on clarifying misunderstandings related to the mathematical transformations involved.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on understanding the reasoning behind each step rather than simply arriving at a solution.

escobar147
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Solve the following first order, ordinary differential equations using separation of variables:

dy/dx = y^2 x

subject to y=-1 when x=0

the correct answer is: y = -2/x^2 + 2

i cannot seem to get this answer, after i separate the variables and integrate both sides i get:

y^2 x^2/2

then if i plug the values in i get y = 0?

please help
 
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When you separate the variables you should get

(1/y^2)dy = x dx

Integrating 1/y^2 wrt y will not give you y^2. Rewrite it as y^-2.
 
rock.freak667 said:
When you separate the variables you should get

(1/y^2)dy = x dx

Integrating 1/y^2 wrt y will not give you y^2. Rewrite it as y^-2.

hi, thanks foryour reply i appreciate it, but could you please explain why when you separate variables y^2 become 1/y^2?
 
dy/dx = y2 x

Divide both sides by y2

(1/y2)dy/dx = x
 
rock.freak667 said:
dy/dx = y2 x

Divide both sides by y2

(1/y2)dy/dx = x

where has the 1 come from? is it the coefficiant of dy/dx?
 
escobar147 said:
where has the 1 come from? is it the coefficiant of dy/dx?

Well you can think of the 1 like that yes.
 

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