Question

x³y⁴+x²y³+xy²+y)dx+(x⁴y³-x³y²-x²y+x)dy=0 find solution​

Answer 1

Answer:

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

or, dy/dx = y.( 1-x^2.y^2)/x.(x^2.y^2+1)…………..(1)

Let x.y= p.

or y= p/x.

dy/dx = ( x.dp/dx - 1.p)/x^2

or, dy/dx = 1/x.dp/dx - p/x^2. putting the value of dy/dx in eqn. (1).

or, 1/x.dp/dx -p/x^2 = (p/x).(1-p^2)/x.(1+p^2).

or, dp/dx - p/x = p.(1-p^2)/x.(1+p^2).

or, dp/dx = p/x.[ (1-p^2)/(1+p^2) + 1].

or, dp/dx = (p/x).[(1-p^2+1+p^2)/(1+p^2)].

or, dp/dx = 2p/x.(1+p^2).

or, {(1+p^2)/p}.dp = (2/x).dx.

or, { 1/p + p}.dp = 2.(1/x).dx.

or, log |p| + (p^2)/2 = 2.log|x| + C.

or, log |x.y| + (1/2).x^2.y^2 = 2.log |x| +C.

or, log|x.y/x^2| + (x^2.y^2)/2 = C.

or, log |y/x| + (x^2.y^2)/2 = C. , Answer.