# Which Method Do I Use?

1. Sep 26, 2006

### Dao Tuat

Could somone help me out and tell me what method to use to solve:

(t^3)(y^2)dt + (t&4)(y^-6)dy = 0

Also, the equation dy/dx - y/x = (x^2)sin2x is homogeneous, right?

Thanks,
Dao Tuat

2. Sep 26, 2006

### dmoravec

for the top one try seperating the variables

3. Sep 26, 2006

### Dao Tuat

Ok, so I seperated them and integrated and ended up with:
y=[7^(6/7)]/(7(ln(t)-c))

Does this look right?

Thanks,
Dao Tuat

Last edited: Sep 27, 2006
4. Sep 27, 2006

### Dao Tuat

Can anyone help me out with the second equation? I'm completly lost.

5. Sep 27, 2006

### dmoravec

that equation for post #3 is not what i got. What did your seperation of variables look like?

EDIT: I'm assuming that equation you have is either $$y=\frac {7^\frac {6}{7}}{7(ln(t)-c)}$$ or $$y=7^\frac{6/7}{7(ln(t)-c))}$$

Last edited: Sep 27, 2006
6. Sep 27, 2006

### Dao Tuat

Yes, it looked like the first one, which is the same as
y=7*sqrt[-1/(7[ln(t)+c])], correct?

What I did was end up with -[(t^3)/(t^4)]dt=[(y^-6)/(y^2)]dy after seperating

I then integrated and got -ln(t)+c=-1/(7y^7), and then solved for y.

For the other equation, dy/dx - y/x = (x^2)sin2x, I got:
y=[(xsin2x)/4]-[([x^2]cos2x)/2]

7. Sep 28, 2006

### Dao Tuat

Do these answeres and approach look right?