# Differential equations

• Jan 1st 2009, 08:25 PM
Kamikazi
Differential equations
Happy new year to all :D

1. dy/dx = (x^2 - y^2) / 2xy
2. x * dy/dx -y = x^3
3. (x+ (e^-x)* siny)dx - (y+ (e^-x) cosx)dy = 0
And which method used to solve these type!
Thanks!(Happy)
Sorry, I don't know how to type math equations in this forum, could anyone show me(Wink)
• Jan 1st 2009, 09:07 PM
mr fantastic
Quote:

Originally Posted by Kamikazi
Happy new year to all :D

1. dy/dx = (x^2 - y^2) / 2xy
2. x * dy/dx -y = x^3
3. (x+ (e^-x)* siny)dx - (y+ (e^-x) cosx)dy = 0
And which method used to solve these type!
Thanks!(Happy)
Sorry, I don't know how to type math equations in this forum, could anyone show me(Wink)

1. $\frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = \frac{1 - \left(\frac{y}{x}\right)^2}{2 \left(\frac{y}{x}\right)}$

The usual technique is to now make the substitution $y = xv$.

2. Re-write as $\frac{dy}{dx} - \frac{1}{x} y = x^2$. Now use the integrating factor method.

Note that the integrating factor is $e^{\int -\frac{1}{x} \, dx} = e^{- \ln |x|} = e^{\ln \frac{1}{|x|}} = \frac{1}{x}$.

3. I don't have time now to check right now but you might need to find an integrating factor to make it exact.
• Jan 1st 2009, 11:18 PM
Kamikazi
Quote:

Originally Posted by mr fantastic
1. $\frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = \frac{1 - \left(\frac{y}{x}\right)^2}{2 \left(\frac{y}{x}\right)}$

The usual technique is to now make the substitution $y = xv$.

2. Re-write as $\frac{dy}{dx} - \frac{1}{x} y = x^2$. Now use the integrating factor method.

Note that the integrating factor is $e^{\int -\frac{1}{x} \, dx} = e^{- \ln |x|} = e^{\ln \frac{1}{|x|}} = \frac{1}{x}$.

3. I don't have time now to check right now but you might need to find an integrating factor to make it exact.

Thanks a lot, now I understand:D
• Jan 2nd 2009, 07:00 AM
Jester
As for the third, it's exact as it stands if the cos x is a cos y. :)