Question

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

Answer 1

\[(3x-y+1)\text d x-(6x-2y-3)\text d y=0\]
\[(3x-y+1)\text d x=(6x-2y-3)\text d y\]
\[\frac{\text d y}{\text d x}=\frac{3x-y+1}{6x-2y-3}\]
\[2\frac{\text d y}{\text d x}=\frac{6x-2y+2}{6x-2y-3}\]
\[\text {let } w=6x-2y\]
\[\frac{\text d w }{\text d x}=6-2\frac{\text d y }{\text dx}\]
\[2\frac{\text d y}{\text d x}=6-\frac{\text d w }{\text d x}\]
\[2\frac{\text d y}{\text d x}=\frac{w+2}{w-3}\]
\[6-\frac{\text d w }{\text d x}=\frac{w+2}{w-3}\]
\[6-\frac {w+2} {w-3}=\frac{\text d w }{\text d x}\]
\[\frac {6(w-3)-w+2} {w-3}=\frac{\text d w }{\text d x}\]
\[\frac {5w-16} {w-3}=\frac{\text d w }{\text d x}\]
\[\text d x=\frac{w-3}{5w-16}\text d w\]
\[\int\text d x=\int\frac{w-3}{5w-16}\text d w\]

Answer 2

that is all i got so far.

Answer 3

\[\ln \text dx =\int\frac{w-3}{5w-16}\text d x\]

Answer 4

Making u = 3x - y maybe easier. If I did it correctly, you should end up with: \[\frac{du}{dx} = \frac{5u-2}{2u-3}\]So the integral becomes: \[ \int \frac{2u -3}{u-2}du = \int 5dx \implies \int (\frac{1}{u-2} + 2)du = 5x + C\]As for your integral, I think both doing long division and partial fractions should solve it.

Answer 5

those are the worst
i will re do this with
u = 3x - y

Answer 6

I agree. I always disliked partial fractions. But it seems that they happen all the time when we are solving DE :-(

Answer 7

takes up to much paper

Answer 8

\[\text {let } w= 3x - y\]\[\frac{\text d w }{\text d x}=3-\frac{\text d y }{\text dx}\]\[\frac{\text d y}{\text d x}=3-\frac{\text d w }{\text d x}\]\[3-\frac{\text d w }{\text d x}=\frac{w+1}{2w-3}\]\[\frac{\text d w }{\text d x}=3-\frac{w+1}{2w-3}\]\[\frac{\text d w }{\text d x}=\frac{3(2w-3)- w+1}{2w-3}\]\[\frac{\text d w }{\text d x}=\frac{5w-8}{2w-3}\]

Answer 9

Stupid me, it should be \(\large \frac{5(u-2)}{2u - 3} \). And, I got: \[\ u + 1 - (-\frac{du}{dx} + 3)(6x - 3 - 2(3x - u)) = 0 \]After substituting in.