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Totally confused on this ODE. Help!

I'm totally stuck on part ii of this question.

Q i) Solve dydt=y3y \frac{dy}{dt} = y^3 - y,
y=12y = \frac{1}{\sqrt2} when t=0 t = 0 .

Answer is:
y=11+e2t y = \frac{1}{\sqrt {1 + e^{2t}}}

Q ii) solve for y again using subsitution y=z12 y = z^{-\frac{1}{2}}, where z=z(t)z = z(t) is a new dependent variable.

This is where I'm totally stuck and I have absolutely no idea how to solve part ii. I know you need to use the chain rule for y=z12y = z^{-\frac{1}{2}}, however, even here I'm unsure what to do with it.

Could someone solve this please?
Reply 1
Original post by BAD AT MATHS
I'm totally stuck on part ii of this question.

Q i) Solve dydt=y3y \frac{dy}{dt} = y^3 - y,
y=12y = \frac{1}{\sqrt2} when t=0 t = 0 .

Answer is:
y=11+e2t y = \frac{1}{\sqrt {1 + e^{2t}}}

Q ii) solve for y again using subsitution y=z12 y = z^{-\frac{1}{2}}, where z=z(t)z = z(t) is a new dependent variable.

This is where I'm totally stuck and I have absolutely no idea how to solve part ii. I know you need to use the chain rule for y=z12y = z^{-\frac{1}{2}}, however, even here I'm unsure what to do with it.

Could someone solve this please?


y=z1/2y=z^{-1/2}

Differentiate to get dydt=12z3/2dzdt\frac{dy}{dt}=-\frac{1}{2}z^{-3/2}\frac{dz}{dt}.

y3=z3/2y^3=z^{-3/2}.

Rewrite the D.E in terms of z and t.
Original post by BAD AT MATHS
I'm totally stuck on part ii of this question.

Q i) Solve dydt=y3y \frac{dy}{dt} = y^3 - y,
y=12y = \frac{1}{\sqrt2} when t=0 t = 0 .

Answer is:
y=11+e2t y = \frac{1}{\sqrt {1 + e^{2t}}}

Q ii) solve for y again using subsitution y=z12 y = z^{-\frac{1}{2}}, where z=z(t)z = z(t) is a new dependent variable.

This is where I'm totally stuck and I have absolutely no idea how to solve part ii. I know you need to use the chain rule for y=z12y = z^{-\frac{1}{2}}, however, even here I'm unsure what to do with it.

Could someone solve this please?


y=z12dydt=12z32dzdty=z^{-\frac{1}{2}} \to \frac{dy}{dt}=-\frac{1}{2}z^{-\frac{3}{2}} \frac{dz}{dt}
Then substitute and separate the variables
Reply 3
Original post by brianeverit
y=z12dydt=12z32dzdty=z^{-\frac{1}{2}} \to \frac{dy}{dt}=-\frac{1}{2}z^{-\frac{3}{2}} \frac{dz}{dt}
Then substitute and separate the variables


Original post by BabyMaths
y=z1/2y=z^{-1/2}

Differentiate to get dydt=12z3/2dzdt\frac{dy}{dt}=-\frac{1}{2}z^{-3/2}\frac{dz}{dt}.

y3=z3/2y^3=z^{-3/2}.

Rewrite the D.E in terms of z and t.


Ok, I see now thanks.

EDIT: babymaths i tried +ve rep'ing you but I got a message saying "Please rate some other members before rating this member again.".
(edited 10 years ago)
Reply 4
Original post by BAD AT MATHS
Ok, I see now thanks.

EDIT: babymaths i tried +ve rep'ing you but I got a message saying "Please rate some other members before rating this member again.".


It really is the thought that counts. Thanks. :smile:

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