integrating factor

Problem 04
$y(4x + y)~dx - 2(x^2 - y)~dy = 0$
 

Solution 04
$M~dx + N~dy = 0$

$y(4x + y)~dx - 2(x^2 - y)~dy = 0$
 

$M = y(4x + y) = 4xy + y^2$

$N = -2(x^2 - y) = -2x^2 + 2y$
 

$\dfrac{\partial M}{\partial y} = 4x + 2y$

$\dfrac{\partial N}{\partial x} = -4x$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (4x + 2y) - (-4x)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 8x + 2y$
 

Problem 03
$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
 

Solution 03
$M~dx + N~dy = 0$

$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
 

$M = y(2x - y + 1) = 2xy - y^2 + y$

$N = x(3x - 4y + 3) = 3x^2 - 4xy + 3x$
 

$\dfrac{\partial M}{\partial y} = 2x - 2y + 1$

$\dfrac{\partial N}{\partial x} = 6x - 4y + 3$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x - 2y + 1) - (6x - 4y + 3)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -4x + 2y - 2$

Problem 02
$2y(x^2 - y + x)~dx + (x^2 - 2y)~dy = 0$
 

Solution 02
$M~dx + N~dy = 0$

$2y(x^2 - y + x)~dx + (x^2 - 2y)~dy = 0$
 

$M = 2y(x^2 - y + x) = 2x^2y - 2y^2 + 2xy$

$N = x^2 - 2y$
 

$\dfrac{\partial M}{\partial y} = 2x^2 - 4y + 2x$

$\dfrac{\partial N}{\partial x} = 2x$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x^2 - 4y + 2x) - 2x$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 2x^2 - 4y = 2(x^2 - 2y)$
 

Problem 01
$(x^2 + y^2 + 1)~dx + x(x - 2y)~dy = 0$
 

Solution 01
$M~dx + N~dy = 0$

$(x^2 + y^2 + 1)~dx + x(x - 2y)~dy = 0$
 

$M = x^2 + y^2 + 1$

$N = x(x - 2y) = x^2 - 2xy$
 

$\dfrac{\partial M}{\partial y} = 2y$

$\dfrac{\partial N}{\partial x} = 2x - 2y$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 2y - (2x - 2y)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -2x + 4y$

Problem 11
$y(x^2 + y^2 - 1) \, dx + x(x^2 + y^2 + 1) \, dy = 0$
 

Solution 11
$y(x^2 + y^2 - 1) \, dx + x(x^2 + y^2 + 1) \, dy = 0$

$y(x^2 + y^2) \, dx - y \, dx + x(x^2 + y^2) \, dy + x \, dy = 0$

$[ \, y(x^2 + y^2) \, dx + x(x^2 + y^2) \, dy \, ] - (y \, dx - x \, dy) = 0$

$(x^2 + y^2)(y \, dx + x \, dy) - (y \, dx - x \, dy) = 0$

$(y \, dx + x \, dy) - \dfrac{y \, dx - x \, dy}{x^2 + y^2} = 0$

$d(xy) - d [ \, \arctan (y/x) \, ] = 0$

$\displaystyle \int d(xy) - \int d[ \, \arctan (y/x) \, ] = 0$

Problem 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

Solution 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

$y^3 \, dx + y \, dx + xy^2 \, dy - x \, dy = 0$

$(xy^2 \, dy + y^3 \, dx) + (y \, dx - x \, dy) = 0$

$y^2(x \, dy + y \, dx) + (y \, dx - x \, dy) = 0$

$(x \, dy + y \, dx) + \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

$d(xy) + d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle \int d(xy) + \int d\left( \dfrac{x}{y} \right) = 0$

$xy + \dfrac{x}{y} = c$

$xy^2 + x = cy$

Problem 05
$y(x^4 - y^2) \, dx + x(x^4 + y^2) \, dy = 0$
 

Problem 05
$y(x^4 - y^2) \, dx + x(x^4 + y^2) \, dy = 0$

$x^4y \, dx - y^3 \, dx + x^5 \, dy + xy^2 \, dy = 0$

$(x^4y \, dx + x^5 \, dy) + (xy^2 \, dy - y^3 \, dx) = 0$

$x^4(y \, dx + x \, dy) + y^2(x \, dy - y \, dx) = 0$

$(y \, dx + x \, dy) + \dfrac{y^2(x \, dy - y \, dx)}{x^4} = 0$

$(y \, dx + x \, dy) + \dfrac{y^2}{x^2} \left( \dfrac{x \, dy - y \, dx}{x^2} \right) = 0$

Problem 04
$2t \, ds + s(2 + s^2t) \, dt = 0$
 

Solution 04
$2t \, ds + s(2 + s^2t) \, dt = 0$

$2t \, ds + 2s \, dt + s^3t \, dt = 0$

$(2t \, ds + 2s \, dt) + s^3t \, dt = 0$

$2(t \, ds + s \, dt) + s^3t \, dt = 0$

$2 \, d(st) + s^3t \, dt = 0$

$\dfrac{2 \, d(st)}{s^3t^3} + \dfrac{s^3t \, dt}{s^3t^3} = 0$

$\dfrac{2 \, d(st)}{(st)^3} + \dfrac{dt}{t^2} = 0$

$2 (st)^{-3} \, d(st) + t^{-2} \, dt = 0$

$\displaystyle 2\int (st)^{-3} \, d(st) + \int t^{-2} \, dt = 0$

$-(st)^{-2} - t^{-1} = -c$