by inspection

Problem 11 | Integrating Factors Found by Inspection

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 - 07 | Integrating Factors Found by Inspection

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 | Integrating Factors Found by Inspection

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 | Integrating Factors Found by Inspection

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

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

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

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

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

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

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

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

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

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

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

Problem 03 | Integrating Factors Found by Inspection

Problem 03
$(x^3y^3 + 1) \, dx + x^4y^2 \, dy = 0$
 

Solution 03
$(x^3y^3 + 1) \, dx + x^4y^2 \, dy = 0$

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

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

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

$x^3y^2 \, d(xy) + dx = 0$
 

Divide by x both sides
$x^2y^2 \, d(xy) + \dfrac{dx}{x} = 0$

$\displaystyle \int (xy)^2 \, d(xy) + \int \dfrac{dx}{x} = 0$

$\frac{1}{3}(xy)^3 + \ln x = -\ln c$

$x^3y^3 + 3\ln x = -3\ln c$

$x^3y^3 = -3\ln c - 3\ln x$

$x^3y^3 = -3(\ln c + \ln x)$

Problem 02 | Integrating Factors Found by Inspection

Problem 02
$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$
 

Solution 02
$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$

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

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

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

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

Divide by y3
$d(xy) - \dfrac{x}{y} \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

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

 
 
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