Help on Differential Equations Please

I need help on solving the following:

1.) (xy)dx + (x^2 - 3y)dy = 0
2.) ( k(e^(2v)) - u ) du = 2( e^(2v) )( (e^(2v)) + ku ) dv

NOTE: item number 2 looks quite confusing, e is the constant with value 2.71828183...
also, take note of the parentheses, I already carefully grouped the expressions, the e's are
basically raised to (2v). k is just a constant, hope that helps...

IN ADDITION, our teacher said that these can be solved using the approach used to solved BERNOULLI TYPE EQUATIONS, since that's the case, can you also show how to solve these by applying the technique used to solve Bernoulli's equations. Complete solutions will be very much be appreciated, i.e instant 5 stars : )

For these format question I mostly use exact equation method. In particular the second equation is very easy to solve with exact format method. If you use bernoulli's method these questions can get very messy.

1) (xy)dx + (x^2-3y)dy = 0;
According to exact equation R(sub x)=xy, R(sub y)=(X^2-3y);
R(sub xy)=R(sub yx)
R(sub xy)=x, R(sub yx)=2x
since 2x is not equal to x, you have to use the integrating factor.

Assume the integrating factor is D. and D=D(y) [D depends only on y]
D(xy)dx+ D(x^2-3y) dy=0
Now again do the R sub business;
R(sub xy)=R(sub yx)
R(sub xy)=D'xy+Dx
R(sub yx)=D(2x)
if we equate bith R(sub xy)=R(subyx)
D'xy+Dx=2Dx;
D'xy=Dx;
D'y=D;
dD/dy *y=D;
dD/D=dy/y;
D=y;

D(xy)dx+D(x^2-3y)=0
xy^2 dx+ x^2y-3y^2=0
again R sub......

R(sub x)=xy^2
R(x)=x^2*y^2/2;
since R also depends on y;
R(sub y)= x^2y-3y^2
R=x^2y^2/2-y^3;
according to exact equations R is a constant therefore c=x^2y^2/2-y^3;
You can leave it like this. If you try to solve the cubic of y it will get really messy.

2) For this one exact equations are much easier than the first one because you don't need an integrating factor here.
k(e^(2v)-u)du=2(e^(2v)(e^(2v)+ku))dv
R(sub u)= k(e^(2v)-u)
R(sub v)=2(e^2v(e^(2v)+ku))
R(sub uv)=k(2e^(2v))=2ke^(2v);
R(sub vu)= 2ke^(2v)'
since R(sub uv)=R(sub vu), this is an exact equation.

R(u)=ku*e^(2v) - (ku^2)/2;
since R also depend on v
R(sub v)=2(e^(4v)+2ku*e^(2v))
R(v)=(e^4v)/2 + ku*e^(2v)
R(u,v)= ku*e^(2v) + e^(4v)/2-(ku^2)/2;
according to exact equation R is a constant.
c=ku*e^(2v) + e^(4v)/2-(ku^2)/2;

Is this a high school Diff. eq?
I hope this helps you. With regular text it is very hard to write subscripts and differentials however I tried. Best of luck.