Question

Solve the DE
(x+2y-1)dx- (2x+y-5)dy=0

Answer 1

by inspection?

Answer 2

did u check if it is exact ?

Answer 3

uhmm..it is exact.. but we're (told) supposed to solve it by either inspection, by obvious subs., by formulas, others .. not by the previous lesson like exact :)

Answer 4

please help me @ganeshie8

Answer 5

can you check if the sign sign between differentials is really "-" ?

Answer 6

(x+2y-1)dx\(\color{red}{-}\) (2x+y-5)dy=0

Answer 7

i think that should be "+", can you please double check

Answer 8

wait

Answer 9

it wont be exact if that sign is "-"

Answer 10

it's plus..

Answer 11

i'm gonna try to solve it too

Answer 12

nope...

Answer 13

nope for what ?

Answer 14

it is a minus sign

Answer 15

Oh its okay, so you want to work it by substitution is it ?

Answer 16

here is the same question, let me know if soemthing doesn't make sense in this :
http://math.stackexchange.com/questions/448839/how-to-solve-x-2y-4dx-2x-y-5dy-0

Answer 17

coefficient of linear : intersecting lines

Answer 18

yes try substituting :
\[x = X-a \\ y = Y-b\]

Answer 19

choose \(a\) and \(b\) valuse such that they eliminate the constants in the original equaiton

Answer 20

we use :
x=u+h
y=v+k

Answer 21

anything is fine :)

Answer 22

(h,k)=(3,-1)

Answer 23

perfect !

Answer 24

so
x=y+3
y=v-1
??

Answer 25

i mean x=u+ 3

Answer 26

yes,
x = u + 3
y = v - 1

Answer 27

resultant equation would be :
\[\large (u+2v)du - (2u+v)dv = 0\]


?

Answer 28

yes, mine too :)

Answer 29

and then?

Answer 30

let \(\large v = tu\)
\(\large dv = tdu + udt\)

Answer 31

homogeneous?

Answer 32

\[\large (u+2v)du - (2u+v)dv = 0\]

becomes

\[\large (u+tu)du - (2u+tu)(tdu + udt ) = 0\]

\[\large (1+t)du - (2+t)(tdu + udt ) = 0\]

\[\large (1+t-t(1+t))du - u(2+t) dt = 0\]

\[\large (1-t^2)du - u(2+t) dt = 0\]

\[\large (1-t^2)du = u(2+t) dt \]

\[\large \dfrac{du}{u} = \dfrac{2+t}{1-t^2} dt \]

Answer 33

verify if the calculaitons are correct or not :) if everything is right, you may integrate both sides

Answer 34

\[C= \ln(x-3) + 3/2 \ln(1-\frac{ y+1 }{ x-3 }) - 1/2 \ln(1+ \frac{ y+1 }{ x-3 })\]
???

Answer 35

something like that... just check the signs again ^^

Answer 36

THANKS!!!!