Question

help me solve this DE please.

(2x+y)dx - (3x+2y-2)dy=0

Answer 1

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

Answer 2

im not sure there is a nice solution

Answer 3

have u tried?till which step did u reach,@UnkleRhaukus gave u proper start...
which substitution u have tried??

Answer 4

u can show that the sub \(X=x−x_0\) and \(Y=y−y0\) will change the
nonhomogeneous diff equation\[\frac{dy}{dx}=\frac{ax+by+c}{ex+fy+g}\]to homogeneous diff equation\[\frac{dY}{dX}=\frac{aX+bY}{eX+fY}\]where\( (x_0,y_0)\) is solution of system\[ax+by+c=0\]\[ex+fy+g=0\]

Answer 5

non exact ode .i think integrating factor can be used to make it exact.

Answer 6

and yes sami...also integrating factor will work

Answer 7

ok. i'll try using integrating factor. thank you.

Answer 8

let us know if its workin or not...

Answer 9

do you think bernoulli's equation would help?

Answer 10

emm...i dont think so

Answer 11

it's hard, to get the integrating factor. lots of variables here.

Answer 12

\[\frac{\text dy}{\text dx}=\frac{2x+y}{3x+2y-2}\qquad\Longrightarrow\frac{\text dY}{\text dX}=\frac{2X+Y+2x_0+y_0}{3X+2Y+3x_0+2y_0-2}\]
\[\qquad\qquad\qquad \qquad\qquad\qquad\qquad2x_0+y_0=0\]\[\qquad\qquad\qquad\qquad\qquad\qquad 3x_0+2y_0-2=0\]

Answer 13

\[3x_0+2(-2x_0)-2=0\]\[-x_0-2=0\]\[x_0=-2\]
\[2(-2)+y_0=0\]\[y_0=4\]

Answer 14

\[\frac{\text dY}{\text dX}=\frac{2X+Y}{3X+2Y}\]\[\qquad=\frac{2+Y/X}{3+2Y/X}\]
\[V=Y/X\qquad Y=VX\]
\[\frac{\text d Y}{\text dX}=V+V'X\]
\[V+V'X=\frac{2+V}{3+V}\]
\[V'X=\frac{2+V}{3+V}~-V\]
which is separable

Answer 15

now how am i going to integrate this:

(2z+1)dz / (z^2 - z -1)

Answer 16

\[\frac{2z+1}{z^2 - z -1}~=\frac{2z-1}{z^2 - z -1}~+\frac{2}{z^2-z-1}\]

Answer 17

is this a partial fraction?

Answer 18

the first fraction has the numerator as the derivative for the denominator,

Answer 19

yes of course! thanks a lot!

Answer 20

how come the numerator of the first fraction becaME 2z-1?

Answer 21

@UnkleRhaukus

Answer 22

well we had to add 2 to get like that

Answer 23

\[\int\frac{2z+1}{z^2 - z -1}~~\text dz ~=\int\frac{2z-1}{z^2 - z -1}~~\text dz+\int\frac{2}{z^2-z-1}~~\text dz\]

Answer 24

the first intergal is easy now
\[u= z^2-z-1\]\[\text du=(2z-1) \text dz\]

\[=\int\frac{\text du}{u}+\int\frac{2}{z^2-z-1}~~\text dz\]

Answer 25

so how about the next fraction? how am i going to integrate it?

Answer 26

firstly complete the square for denum

Answer 27

how about -5/4 after completing the square, where does it go?

Answer 28

try to use this :\[\int \frac{u'}{u^2-a^2}dx = \frac{1}{2a}Log\left | \frac{u-a}{u+a} \right | + C\]

Answer 29

sorry
\[\int \frac{u'}{u^2-a^2}dx = \frac{1}{2a}\ln\left | \frac{u-a}{u+a} \right | + C\]

Answer 30

ok. thank you. i have another DE to solve. help me again please.

Answer 31

np post it...