How to Solve a Differential Equation Using Separation of Variables

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The discussion focuses on solving the differential equation x^2dy/dx = y - xy using separation of variables, with the initial condition y(-1) = -1. After correctly separating and integrating, the user arrived at 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 to resolving the discrepancy lies in algebraic manipulation, specifically recognizing that e^(-ln x) simplifies to 1/x. By applying this transformation, the user can derive the book's solution and determine the constant c. The conversation emphasizes the importance of algebra in solving differential equations after integration.
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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...
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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