Question

\[1^{st}\text{ order ODE }\\ (xy+1)dx+(2y-x)dy=0\]

Answer 1

Working with exact equations?

Answer 2

yes but this one is not exact and the intergration factor is ugly

Answer 3

\[M(x,y)=xy+1~~\Rightarrow~~\frac{\partial M}{\partial y}=x\\
N(x,y)=2y-x~~\Rightarrow~~\frac{\partial N}{\partial x}=-1\]
Our integrating factor can be either
\[\exp\left(\int\frac{M_y-N_x}{N}~dx\right)~~\text{or}~~\exp\left(\int\frac{N_x-M_y}{M}~dy\right)\]
\[\int\frac{x+1}{2y-x}~dx~~\text{or}~~-\int\frac{x+1}{xy+1}~dy\]
The second integral looks like it might be easier to work with, so I'd try that.

Answer 4

but our proffesor told us we shud not do intergration factor with two variables...like this one,....

Answer 5

so i was thinking maybe another way cud do cos we didnt learnt multivariate function

Answer 6

is it seperable with a y=vx substitution?

Answer 7

when i did that substitution it gives,
\[\large (vx^2+2v^2x-vx+1)dx+x^2(2v-1)dv=0\]
still not exact,dint c hw this cn be seperable

Answer 8

very intrasting one i'm also trying .........