How to Solve a Differential Equation Using Separation of Variables

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SUMMARY

The discussion focuses on solving the differential equation x²(dy/dx) = y - xy using the method of separation of variables, with the initial condition y(-1) = -1. The user initially derived the solution y = e^(-1/x - ln x + C) but struggled to match it with the book's answer, y = e^(-(1 + 1/x))/x. The key insight provided was the algebraic manipulation of the exponential function, specifically recognizing that e^(-ln x) simplifies to 1/x, leading to the correct form of the solution.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Proficiency in calculus, particularly integration techniques.
  • Familiarity with the method of separation of variables in solving differential equations.
  • Basic knowledge of exponential functions and logarithmic identities.
NEXT STEPS
  • Review the method of separation of variables in differential equations.
  • Study algebraic manipulation of exponential and logarithmic functions.
  • Practice solving first-order differential equations with initial conditions.
  • Explore additional examples of differential equations to reinforce understanding.
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and differential equations, as well as anyone seeking to improve their problem-solving skills in this area.

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
\displaystyle e^{-1/x-\ln(x)+C}=e^{-1/x}e^{-\ln|x|}e^C
\displaystyle =e^{-1/x}(1/x)e^C​
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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