Problem 02 | Substitution Suggested by the Equation

$\sin y(x + \sin y)~dx + 2x^2 \cos y~dy = 0$
 

Let
$z = \sin y$

$dz = \cos y~dy$
 

Hence,
$z(x + z)~dx + 2x^2~dz = 0$       → homogeneous equation
 

Let
$z = vx$

$dz = v~dx + x~dv$
 

$vx(x + vx)~dx + 2x^2(v~dx + x~dv) = 0$

$x^2(v + v^2)~dx + 2vx^2~dx + 2x^3~dv = 0$

$[ \, x^2(v + v^2)~dx + 2vx^2~dx \, ] + 2x^3~dv = 0$

$x^2 [ \, (v + v^2) + 2v \, ]~dx + 2x^3~dv = 0$

$x^2(3v + v^2)~dx + 2x^3~dv = 0$

$vx^2(3 + v)~dx + 2x^3~dv = 0$
 

Divide by vx³(3 + v)
$\dfrac{dx}{x} + \dfrac{2~dv}{v(3 + v)} = 0$

$\dfrac{dx}{x} + \dfrac{2~dv}{v(3 + v)} = 0$
 

Consider
$\dfrac{2}{v(3 + v)} = \dfrac{A}{v} + \dfrac{B}{3 + v}$

$2 = A(3 + v) + Bv$
 

Set v = 0, A = 2/3
Set v = -3, B = -2/3
 

$\dfrac{2}{v(3 + v)} = \dfrac{2/3}{v} - \dfrac{2/3}{3 + v}$
 

Thus,
$\dfrac{dx}{x} + \left(\dfrac{2/3}{v} - \dfrac{2/3}{3 + v} \right)~dv = 0$

$\displaystyle \int \dfrac{dx}{x} + \dfrac{2}{3}\int \dfrac{dv}{v} - \dfrac{2}{3}\int \dfrac{dv}{3 + v} = 0$

$\ln x + \frac{2}{3}\ln v - \frac{2}{3}\ln (3 + v) = \frac{1}{3}\ln c$

$3\ln x + 2\ln v - 2\ln (3 + v) = \ln c$

$\ln x^3 + \ln v^2 - \ln (3 + v)^2 = \ln c$

$\ln \dfrac{x^3v^2}{(3 + v)^2} = \ln c$

$\dfrac{x^3v^2}{(3 + v)^2} = c$

$x^3\left( \dfrac{v}{3 + v} \right)^2 = c$
 

From
$z = vx$

$v = \dfrac{z}{x}$
 

$x^3\left( \dfrac{\dfrac{z}{x}}{3 + \dfrac{z}{x}} \right)^2 = c$

$x^3\left( \dfrac{\dfrac{z}{x}}{\dfrac{3x + z}{x}} \right)^2 = c$

$x^3\left( \dfrac{z}{3x + z} \right)^2 = c$
 

But   $z = \sin y$
 

$x^3\left( \dfrac{\sin y}{3x + \sin y} \right)^2 = c$

$x^3 \cdot \dfrac{\sin^2 y}{(3x + \sin y)^2} = c$

$x^3 \sin^2 y = c(3x + \sin y)^2$           answer