Differential Equations

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August 30th 2008, 05:41 AM
Tangera

Differential Equations

Solve the following differential equation using the substitution given:

$2x \displaystyle{\frac{dy}{dx}} = 2y - y^2$ , by substituting y = zx.

I got to a stage where zx = 2 (not sure if I am right)...and then I don't know how to continue. Please help! Thank you!
August 30th 2008, 05:48 AM
Moo

Hello,

Quote:

Originally Posted by Tangera

Solve the following differential equation using the substitution given:

$2x \displaystyle{\frac{dy}{dx}} = 2y - y^2$ , by substituting y = zx.

I got to a stage where zx = 2 (not sure if I am right)...and then I don't know how to continue. Please help! Thank you!

Don't forget that z is a function of x.

$\frac{dy}{dx}=z+x \frac{dz}{dx}$

You must arrive to $2 \frac{dz}{dx}+z^2=0$
August 30th 2008, 05:52 AM
Tangera

Quote:

Originally Posted by Moo

Hello,

Don't forget that z is a function of x.

$\frac{dy}{dx}=z+x \frac{dz}{dx}$

You must arrive to $2 \frac{dz}{dx}+z^2=0$

Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x
August 30th 2008, 06:00 AM
Moo

Quote:

Originally Posted by Tangera

Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x

Oops sorry ^^

Then, you know that y=xz... and you substitute (Surprised)
August 30th 2008, 06:11 AM
Tangera

Quote:

Originally Posted by Moo

Oops sorry ^^

Then, you know that y=xz... and you substitute (Surprised)

Hmm...just to check...do I substitute y = xz = 2 into the given differential equation? But if so, then I get $2x\displaystyle{\frac{dy}{dx}}$ = 0...and I am lost! >.<
August 30th 2008, 06:16 AM
Moo

Quote:

Originally Posted by Tangera

Hmm...just to check...do I substitute y = xz = 2 into the given differential equation? But if so, then I get $2x\displaystyle{\frac{dy}{dx}}$ = 0...and I am lost! >.<

Yes, and $2(2)-(2)^2=0$ too (Wondering) (Worried)
August 30th 2008, 06:32 AM
Tangera

Quote:

Originally Posted by Moo

Yes, and $2(2)-(2)^2=0$ too (Wondering) (Worried)

The answer I was given is 2x/(x+c)....but I don't know how to get there...Thanks for the help!
August 30th 2008, 06:42 AM
Moo

Haha I know why

Quote:

Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x

The exact solution shall be $x+c=\frac 2z$ (c is the constant of integration !)

$\implies z=\frac{2}{x+c}$

Since $xz=y ~,~ y=\frac{2x}{x+c}$
August 30th 2008, 06:49 AM
Tangera

Ahhh I see!! :D (Clapping)

Thank you very much, Moo!