How to Solve a Differential Equation Using Separation of Variables

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Homework Help Overview

The discussion revolves around solving a differential equation using the method of separation of variables. The specific equation is x^2 dy/dx = y - xy, accompanied by an initial condition of y(-1) = -1.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the separation and integration steps taken to solve the differential equation. There is a focus on understanding the algebraic manipulation needed to reconcile the original solution with the book's answer. Questions arise regarding the transformation of exponential functions and the implications of the initial condition.

Discussion Status

Some participants have provided hints regarding algebraic simplifications and the properties of exponential functions. There is an ongoing exploration of how to derive the book's answer from the participant's initial solution, with no explicit consensus reached yet.

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. The initial condition is a critical aspect of the problem being discussed.

bdh2991
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Homework Statement



use separation of variables to solve the differential equation x^2dy/dx=y-xy

with the initial condition of y(-1)=-1

Homework Equations


The Attempt at a Solution



after i separated and integrated i got the answer y=e^(-1/x-lnx+c)

the answer in the book is y=e^-(1+1/x)/x

i can't figure out how they got to that even after i plugged in -1 for x and y

some help would be greatly appreciated
 
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You've done the calculus correctly, now you need to do a little algebra.

Hint: what's e^(-ln x)?
 
bdh2991 said:

Homework Statement



use separation of variables to solve the differential equation x^2dy/dx=y-xy

with the initial condition of y(-1)=-1

The Attempt at a Solution



after i separated and integrated i got the answer y=e^(-1/x-lnx+c)

the answer in the book is y=e^-(1+1/x)/x

i can't figure out how they got to that even after i plugged in -1 for x and y

some help would be greatly appreciated
[itex]\displaystyle e^{-1/x-\ln(x)+C}=e^{-1/x}e^{-\ln|x|}e^C[/itex]
[itex]\displaystyle =e^{-1/x}(1/x)e^C[/itex]​
Does that help?
 
well i think it would be x^-1 so then that would give me the x in the denominator then plugging in the initial values i should get c=1?
 
bdh2991 said:
well i think it would be x^-1 so then that would give me the x in the denominator then plugging in the initial values i should get c=1?

e0 = 1
 
ok thanks for the help i didn't remember that you could rewrite the e functions that way...
 

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