# I I need help with differential equation(3x^2-2y^2)(dy/dx)=2xy

1. Aug 15, 2016

### edgarpokemon

2. Aug 15, 2016

### haruspex

What about the constant of integration?

3. Aug 15, 2016

### edgarpokemon

if v^2-2=u, then du=2v, which cancels the 2v that is above of the denominator right? help!! how would you do it using the method i used?

4. Aug 15, 2016

### haruspex

2dv.
But you did not answer my question. Go through the integration step and make sure to include a constant of integration. If you still do not see your error, please post all your steps in detail.

5. Aug 15, 2016

### edgarpokemon

I dont see my error!! i will post my steps: (3x^2-2y^2)(dy/dx)=2xy. (3x^2-2y^2)/(2xy)=dx/dy. (3x^2/2xy)-(2y^2/2xy)=dx/dy. 3(x/y)/2-(y/x)=dx/dy. x=yv, v=x/y.
(3/2)v-(1/v)=(dv/dy)y+v. (3/2)v-(1/v)-v=(dv/dy)y. find lcd for left side of equation. lcd is 2v, so (3v^2-2-2v^2)/(2v)=(dv/dy)y. make dv on one side and dy on the other side, so (2v/(v^2-2))dv=dy/y. integral of (2v/(v^2-2))dv: u=v^2-2, du=2vdv. dv=du/2v, the 2v cancels out, leading to a resulting integral of ln|v^2-2|=ln|y|+c. multiply both sides by e leads to, y=(x^2/y^2)-2+c. the values are y=-1 and x=0. so c=y-(x^2/y^2)+2. c=-1+2=1. y=(x^2/y^2)-2+1. y=(x^2/y^2)-1. y=(x^2-y^2)/(y^2).
y^3=x^2-y^2. y^3+y^2=x^2. y^2(y+1)=x^2. but the answer is 2y^2(y+1)=x^2. my question would be the factor of two on the right side. help!!!

6. Aug 15, 2016

### haruspex

No, you do not multiply by e to remove the ln()s. What do you do?
That is wrong.

7. Aug 15, 2016

### edgarpokemon

I thought that it was possible to cancel the natural log by e^x since it is the opposite of the natural log. so will the constant c also be e^c? and I solve for c? help me already!! jaja

8. Aug 15, 2016

### edgarpokemon

oh I got it!!! so ln|y|=ln|c((x^2/y^2)-2). x=0. y=-1. I use e^x on both side of the equation. so c=1/2. then y=1/2((x^2/y^2)-2). then y=(x^2-2y^2)/(2y^2). finally 2y^3+2y^2=x^2. correct? I have never worked a problem were I have to found what c is, so maybe that was my problem?

9. Aug 16, 2016

### haruspex

Where ln(|c|) is the constant of integration, yes.

10. Aug 16, 2016

### edgarpokemon

thank you very much :P