Question

\[(x-2y+1)\text dx+(4x-3y-6)\text dy=0\]
\[\frac{\text dy}{\text dx}=\frac{2y-x-1}{4x-3y-6}\]

Answer 1

\[\frac{\text d X}{ \text d Y}=\frac{2Y+2k-X-h-1}{4X+4h-3Y-3k-6}\]\[[\frac{\text d X}{ \text d Y}=\frac{2Y-X+(2k-h-1)}{4X-3Y+(4h-3k-6)}\]\[(2k-h-1)=0;\qquad(4h-3k-6)=0\]\[\frac{\text d Y}{\text d X}=\frac{2Y-X}{4X-3Y}=\frac{2(Y/X)-1}{4-3(Y/X)}\]
\[V=(Y/X);\qquad Y=VX\]
\[\frac{\text d Y}{\text d X}=X\frac{\text d V}{\text d X}\]
\[X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V\]
\[X\frac{\text d V}{\text d X}=\frac{2V-1}{4-3V}-V=\frac{2V-1}{4-3V}-V\frac{4-3V}{4-3V}\]\[X\frac{\text d V}{\text d X}=\frac{3V^2-2V-1}{4-3V}\]
\[\frac{\text dX}{X}=\frac{4-3V}{3V^2-2V-1}\text dV\]
\[\ln X=\int \frac{1}{4(V-1)}-\frac{15}{4(3V+1)}\text d V \]
\[=\frac{1}{4}\int\frac{1}{V-1}-\frac{15}{3V+1}\text d V\]
\[4\ln(X)=\ln(V-1)-5\ln(3V+1)+c\]\[4\ln(X)=\ln(\frac{Y}{X}-1)-5\ln(3\frac{Y}{X}+1)+c\]\[4\ln(x+3)=\ln\left(\frac{y+2}{x+3}-1\right)-5\ln\left(3\frac{y+2}{x+3}+1\right)+c\]\[(x+3)^4={\frac{y+2}{x+3}-1}+\left(3\frac{y+2}{x+3}+1\right)^{-5} +e^c \]

Answer 2

? did i make a mistake on the last step or somewherelse

Answer 3

Pretty sure you have to use product rule in \(\frac{\text{dY}}{\text{dX}}\).

Answer 4

Thanks, i should have
\[\frac{\text d Y}{\text d X}=X\frac{\text d V}{\text d X}+V\]
right?

Answer 5

i haven't written up the partial fractions bit but i think i have got that right

Answer 6

Yeah, that's about right.

Answer 7

the last step where i raised everything the the base e to get rid of the logs,
i was a bit confused about what to do with the minus in front of the 5

Answer 8

Wait, lemme solve this one.

Answer 9

Simplify the 2nd to last line before raising everything to e, then everything will be clear.

Answer 10

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

Answer 11

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

Answer 12

\[X^4=\frac{V-1}{(3V+1)^5}+c\]

Answer 13

You're supposed to multiply by e^c, not add.

Answer 14

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

Answer 15

\[e^{\text{blabla}+c}=e^{\text{blabla}}e^c\]

Answer 16

your saying my last three lines should be
\[\dots\times e^c\]

Answer 17

Yep. Then replace \(e^c\) with \(A\) to have one constant.

Answer 18

\[(x+3)^4=\frac{\frac{x+3}{y+2}-1}{(3\frac{x+3}{y+2}+1)^5}\times e^c\]
\[(x+3)^4=A\frac{\frac{x+3}{y+2}-1}{(3\frac{x+3}{y+2}+1)^5}\]

Answer 19

\[\left(3 \frac{x+3}{y+2}+1\right)^5 \times (x+3)^4=A\left( \frac{x+3}{y+2}-1\right)\]

Answer 20

is there anything else i can do ,
thank you so much for you help by the way

Answer 21

Not really. You can't isolate y+2. Leave it at that.

Answer 22

Thanks blockcolder
cool , now i just have to do 10,11, and 12