Question

first order ODE
\[(3x-y+1)\text d x-(6x-2y-3)\text d y=0\]

Answer 1

\[(3x-y+1)\text d x=(6x-2y-3)\text d y\]\[\frac{\text d y}{\text d x}=\frac{3x-y+1}{6x-2y-3}\]\[2\frac{\text d y}{\text d x}=\frac{6x-2y+2}{6x-2y-3}\]\[\text {let } w=6x-2y\]\[\frac{\text d w }{\text d x}=6-2\frac{\text d y }{\text dx}\]\[2\frac{\text d y}{\text d x}=6-\frac{\text d w }{\text d x}\]\[2\frac{\text d y}{\text d x}=\frac{w+2}{w-3}\]\[6-\frac{\text d w }{\text d x}=\frac{w+2}{w-3}\]\[6-\frac {w+2} {w-3}=\frac{\text d w }{\text d x}\]\[\frac {6(w-3)-w+2} {w-3}=\frac{\text d w }{\text d x}\]\[\frac {5w-16} {w-3}=\frac{\text d w }{\text d x}\]\[\text d x=\frac{w-3}{5w-16}\text d w\]\[\int\text d x=\int\frac{w-3}{5w-16}\text d w\]

Answer 2

am i doing this right?

Answer 3

true or false thou?

Answer 4

good question