First order differential equation with substitution

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
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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