# Differential Equations

• Aug 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!
• Aug 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 $2 \frac{dz}{dx}+z^2=0$
• Aug 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
• Aug 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)
• Aug 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! >.<
• Aug 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)
• Aug 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!
• Aug 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 $xz=y ~,~ y=\frac{2x}{x+c}$
• Aug 30th 2008, 06:49 AM
Tangera
Ahhh I see!! :D (Clapping)

Thank you very much, Moo!