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ganeshie8 · Answer

@lucy4104

anonymous · Answer

hi ^^

anonymous · Answer

I see you're getting a lesson from the great ganeshie8!

anonymous · Answer

yeshhhhh naw u need to capitalize dat ~ "the Great Ganeshie8" lowercase words are not of appropriate level >3<

ganeshie8 · Answer

$\large   (4-x)dy +2ydx=0 $

$\large   (4-x)dy  =  -2ydx=0 $

$\large   \frac{1}{y}dy  =  \frac{2}{x-4}dx $

Integrate both sides


$\large   \int   \frac{1}{y}dy  =  \int  \frac{2}{x-4}dx $

$\large   \ln |y|     =  2 \ln | x-4 | + \ln  C $


$\large  y     =  C( x-4 )^2 $

ganeshie8 · Answer

lol, did u get above answer  ?

anonymous · Answer

yeah...

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=%284-x%29dy+%2B2ydx%3D0+

anonymous · Answer

Why did he take off ponits???? omg need to tell him

ganeshie8 · Answer

wolfram also gives same answer... yeah u should get full marks...

anonymous · Answer

and then xy^2dy-(x^3+y^3)dx=0 ?

ganeshie8 · Answer

$\large     xy^2dy = (x^3+y^3)dx$

$\large     \frac{dy}{dx} = \frac{x^3+y^3}{xy^2}$

ganeshie8 · Answer

this is clearly a homogeneous equation,  putting x = kx, y = ky will not change the equaiton.

ganeshie8 · Answer

so try the homogeneous substitution :

$\large   y =  vx$

$\large  y' =     v + v'x$

the equaiton b ecomes :

$\large     v + v'x = \frac{x^3+(vx)^3}{x(vx)^2}$

$\large     v + v'x = \frac{x^3(1+v^3)}{x^3v^2}$

$\large     v + v'x = \frac{1+v^3}{v^2}$

ganeshie8 · Answer

fine, so far ?

anonymous · Answer

uhm hahah we weren't supposed to use homogenous, bernoulli, nor ...uhm rthere was something else, but...yeah, sure, y not?

ganeshie8 · Answer

so try the homogeneous substitution :

$\large   y =  vx$

$\large  y' =     v + v'x$

the equaiton b ecomes :

$\large     v + v'x = \frac{x^3+(vx)^3}{x(vx)^2}$

$\large     v + v'x = \frac{x^3(1+v^3)}{x^3v^2}$

$\large     v + v'x = \frac{1+v^3}{v^2}$

$\large     v'x = \frac{1+v^3}{v^2} - v$

$\large     v'x = \frac{1}{v^2}$

$\large     \frac{dv}{dx}x = \frac{1}{v^2}$

$\large     v^2~dv = \frac{1}{x} dx$

$\large     \int v^2~dv = \int \frac{1}{x} dx$

$\large     \frac{v^3}{3} = \ln |x| + C$

ganeshie8 · Answer

substitute back v = y/x above^

ganeshie8 · Answer

the solution is :

$\large \frac{y^3}{3x^3} =   \ln |x| + C$

ganeshie8 · Answer

the solution is :

$\large y^3 =   3x^3   (\ln |x| + C)$

ganeshie8 · Answer

homogeneous is the clean way, im not sure how else we can solve this...

anonymous · Answer

hmmm need to ask my teacher if he's grading these things right...I got the right answer here too...okay, cool thanks for your enormous help~ once, again!! haha need to go grab 3 hours of sleep real fast, and then it's off to school^^ cya later this week, perhaps XD

ganeshie8 · Answer

good night ! have horrible calcAB dreams ;)