Differential Equations

**Tangera** · August 30th 2008, 05:41 AM

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!

**Moo** · August 30th 2008, 05:48 AM

Hello,

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 $2 \frac{dz}{dx}+z^2=0$

**Tangera** · August 30th 2008, 05:52 AM

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

**Moo** · August 30th 2008, 06:00 AM

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

**Tangera** · August 30th 2008, 06:11 AM

Originally Posted by Moo

Oops sorry ^^

Then, you know that y=xz... and you substitute

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! >.<

**Moo** · August 30th 2008, 06:16 AM

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! >.<