Question

\[(x−2y+1)dx+(4x−3y−6)dy=0\]

Answer 1

\[\frac{\text dy}{\text dx}=\frac{2y-x-1}{4x-3y-6}\]

Answer 2

\[ \frac{dy}{dx} = \frac{(x−2y+1)}{(4x−3y−6)}\]
Let, X = x+h, and Y = y+k
\[ \frac{dY}{dX} = \frac{(X−2Y)+ (h - 2k + 1)}{((4X−3Y)+(4h-3y−6)}\]

Now,
h - 2k + 1 = 0
4h - 3k - 6 = 0
http://www.wolframalpha.com/input/?i=solve++h+-+2k+%2B+1+%3D+0%2C+4h+-+3k+-+6+%3D+0

Solve the above homogeneous differential equation above in X and Y, and put the values X and Y back to x and y from h and k

Answer 3

your missing a negative on your first line

Answer 4

Oh .. sorry about that!!

Answer 5

for this kind of questions, first find the solution for the equations in numerator and denominator
then shift the origin to that point
now use y=zx
it becomes easier to solve

Answer 6

@experimentX 's method is right :)

Answer 7

reducible to homogeneous form!!!

Answer 8

\[\frac{\text d X}{ \text d Y}=\frac{2Y+2k-X-h-1}{4X+4h-3Y-3k-6}\]\[=\frac{2Y-X+(2k-h-1)}{4X-3Y+(4h-3k-6)}\]\[(2k-h-1)=0;\qquad(4h-3k-6)=0\]

Answer 9

\[ \frac{dY}{dX} = \frac{2Y - X}{4X - 3Y}\]

let Y = vX <--- usual method

Answer 10

\[k=2\]\[h=3\]

Answer 11

http://www.wolframalpha.com/input/?i=f%27+%3D+%282y+-+x%29%2F%284x+-+3y%29

Answer 12

\[\frac{\text d Y}{\text d X}=\frac{2Y-X}{4X-3Y}=\frac{2(Y/X)-1}{4-3(Y/X)}\]
\[V=(Y/X);\qquad Y=VX\]\[\frac{\text d Y}{\text d X}=X\frac{\text d V}{\text d X}+V\]

Answer 13

\[X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V=\frac{2V-1}{4-3V}-V\frac{4-3V}{4-3V}\]\[X\frac{\text d V}{\text d X}=\frac{3V^2-2V-1}{4-3V}\]

Answer 14

\[\frac{\text d X}{X}=\frac{4-3V}{3V^2-2V-1}\text d V \]

Answer 15

what is the next step , unless i have errors so far

Answer 16

\[ \int \frac{1}{X} dX = \ln X\]

Answer 17

I think we can solve the other guy using partial fraction
http://www.wolframalpha.com/input/?i=partial+fraction+%284-3x%29%2F%283x^2+-+2x+-+1%29

Answer 18

\[\ln X=\int \frac{1}{4(V-1)}-\frac{15}{4(3V+1)}\text d V \]
\[=\frac{1}{4}\int\frac{1}{(V-1)}-\frac{15}{(3V+1)}\text d V\]

Answer 19

\[ \frac{1}{4} \ln (v-1) - 5 \ln(2V + 1)\]

Answer 20

2V or 3V?

Answer 21

\[ \ln X = \frac{1}{4} \ln (V-1) - 5 \ln(3V + 1) + C\]
Putting back values of V,
\[ \ln X = \frac{1}{4} \ln (Y/X-1) - 5 \ln(3Y/X + 1) + C\]

put values of X and Y
X = x + 3
Y = y + 2
then you have answer.

Answer 22

\[4\ln(X+3)=\ln\left(\frac{y+2}{x+3}-1\right)-5\ln\left(3\frac{y+2}{x+3}+1\right)+c\]

\[(X+3)^4={\frac{y+2}{x+3}-1}+\left(3\frac{y+2}{x+3}+1\right)^{-5} +e^c \]?

Answer 23

X*x

Answer 24

\[(x+3)^4={\frac{y+2}{x+3}-1}+\left(3\frac{y+2}{x+3}+1\right)^{-5} +e^c \]is that the answer to my question/

Answer 25

http://www.wolframalpha.com/input/?i=f%27+%3D+-%28x%E2%88%922y%2B1%29%2F%284x%E2%88%923y%E2%88%926%29
looks like we got error

Answer 26

ah bother