Solve by inspection...

Question

( x³ + x y² −y ) d x + ( y³ + x² y + x ) d y = 0

Answer 1

These type of differential equations are simple enough to enable us to find integrating factors

by inspection . The ability to do this depends largely upon recognition of certain common exact differentials.

Here are some exact differentials that occur frequently .

1. d (x y) = x d y + y dx
2. d( x/ y) = [ y d x -x d y ] / y²
3. d( y/ x) = [ x d y -y d x] / x²
4. d [ln [ x / y ] ] = [ y d x -x d y ] / x y
5. d { tan^-1 [ y / x ] } = [ x d y -y d x] / [x² +y²]

Lets get back to our equation. ( x³ + x y² −y ) d x + ( y³ + x² y + x ) d y = 0

or equivalently x³ d x+ x y² d x −y d x + y³ d y+ x² y d y+ x d y =0

Let us group the terms of like degree, writing the equation in the form

x³ d x + x² y d y + y³ d y + x y² d x + x d y −y d x = 0

x² [ x d x + y d y] + y² [ y d y + x d x ] + x d y −y d x = 0

[ x d x + y d y] [ x² + y² ] + x d y −y d x = 0

x d x + y d y + [ x d y −y d x ] / ( x² + y² ) = 0

(1 /2 ) d [ x² ] + (1 /2 ) d [ y² ] + d { tan^-1 [ y / x ] } = 0

(1 /2 ) [ x² ] + (1 /2 ) [ y² ] + { tan^-1 [ y / x ] } = c

x² + y² + 2 tan^-1 [ y / x ] = c

2 tan^-1 [ y / x ] = c -x² - y²

Answer 2

Hi,

( x³ + x y² −y ) d x + ( y³ + x² y + x ) d y = 0

watch: ( x³ + x y² −y ) = x(x^2 + y^2) -y AND ( y³ + x² y + x ) = y(x^2 + y^2) +x

now ( x³ + x y² −y ) d x + ( y³ + x² y + x ) d y = 0 must be divided by (x2 + y^2)

you will have xdx -y/(x2 + y^2)dx + ydy + x/(x2 + y^2) dy =0

Now : xdx + ydy + x/(x2 + y^2) dy -y/(x2 + y^2)dx =0

xdx + ydy + (xdy -ydx) / (x2 + y^2) =0 now any item is familiar for you

1/2x^2 + 1/2y^2 + arctan y/x =0

I hope it is useful,

Minoo