1. ## Differential Equations

Solve the following differential equation using the substitution given:

$\displaystyle 2x \displaystyle{\frac{dy}{dx}} = 2y - y^2$, 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!

2. Hello,
Originally Posted by Tangera
Solve the following differential equation using the substitution given:

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

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

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

3. Originally Posted by Moo
Hello,

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

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

You must arrive to $\displaystyle 2 \frac{dz}{dx}+z^2=0$
Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x

4. 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

5. 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 $\displaystyle 2x\displaystyle{\frac{dy}{dx}}$ = 0...and I am lost! >.<

6. 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 $\displaystyle 2x\displaystyle{\frac{dy}{dx}}$ = 0...and I am lost! >.<
Yes, and $\displaystyle 2(2)-(2)^2=0$ too

7. Originally Posted by Moo
Yes, and $\displaystyle 2(2)-(2)^2=0$ too
The answer I was given is 2x/(x+c)....but I don't know how to get there...Thanks for the help!

8. Haha I know why

Yes I reached there and got xz = 2 after intergrating. And then I am stuck. X.x
The exact solution shall be $\displaystyle x+c=\frac 2z$ (c is the constant of integration !)

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

Since $\displaystyle xz=y ~,~ y=\frac{2x}{x+c}$

9. Ahhh I see!!

Thank you very much, Moo!