Question

If (x + y) dx + (x – y) dy = 0, then
(a) x^2 + 2xy - y^2=C (b) x^2 - 2xy - y^2=C (c) x^2 + 2xy + y^2=C (d) x^2 - 2xy + y^2=C

Answer 1

I solved and got the answer not one of your choices, mine is \(x^2+4xy -y^2 =c\) while your a has just 2 xy. So, don't know

Answer 2

\[(x+y)+(x-y)y'=0\]
If \(M(x,y)=x+y\) and \(N(x,y)=x-y\), check if the equation is exact:
\[\frac{\partial M}{\partial y}=1~~~~~~~~~~\frac{\partial N}{\partial x}=1\]
The partial derivatives are equivalent, so you have an exact equation.

So now you have \(\Psi_x=M(x,y)=x+y.\)
\[\Psi_x=x+y\\
\int\Psi_x~dx=\int(x+y)~dx\\
\Psi=\frac{1}{2}x^2+xy+h(y)\]
Differentiate \(\Psi\) w.r.t to y:
\[\Psi_y=N(x,y)\\
x+h'(y)=x-y\\
h'(y)=-y\\
h(y)=-\frac{1}{2}y^2+C\]
So the solution is \(\Psi=C\):
\[\frac{1}{2}x^2+xy-\frac{1}{2}y^2=C\\
x^2+2xy-y^2=C\]