First order differential equation with substitution

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SUMMARY

The discussion focuses on solving the first-order differential equation \( t^2 y' + 4ty - y^3 = 0 \) using the substitution \( v = y^{-2} \). The user initially attempted to isolate \( y' \) and manipulate the equation but struggled to separate variables. A key insight was provided by another participant, suggesting the use of an integrating factor after recognizing the potential to divide by \( y^3 \) instead of \( t^2 \). This approach simplifies the problem and leads to a more straightforward solution.

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kaitamasaki
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Homework Statement



t^2 y' + 4ty - y^3 = 0

Homework Equations



Hint was given in the question: substitute with v = y^-2

The Attempt at a Solution



Dividing by t^2 and isolating y':
t^2 y' = y^3 - 4ty
y' = y^3 / t^2 - 4y/t

dv/dt = 0
y = v^(-1/2)
dy/dt = (-1/2)v^(-3/2) v'

so y' = dy/dt = (-1/2)v^(-3/2) v' = v^(-3/2) - 4t(v^(-1/2))

But after playing with algebra I cannot separate v and t,
I end up with;
dv/dt - 8v/t = -2/t^2
Where v' = dv/dt

How should I have approached this problem?
 
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hi kaitamasaki! :smile:

(try using the X2 icon just above the Reply box :wink:)
kaitamasaki said:
Dividing by t^2 …

why t2 ? :confused:

try dividing by y3 :smile:
 
Oh gee can't believe I missed that
Had to use integrating factor after
 
Last edited:

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