# Thread: Differential equations

1. ## Differential equations

Happy new year to all
I 've just studying differential equations and I 've some equations need solve, please help me!

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!
Sorry, I don't know how to type math equations in this forum, could anyone show me

2. Originally Posted by Kamikazi
Happy new year to all
I 've just studying differential equations and I 've some equations need solve, please help me!

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!
Sorry, I don't know how to type math equations in this forum, could anyone show me
1. $\displaystyle \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = \frac{1 - \left(\frac{y}{x}\right)^2}{2 \left(\frac{y}{x}\right)}$

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

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

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

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

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

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

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

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

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

4. As for the third, it's exact as it stands if the cos x is a cos y.