Question -3

This can be also done,

$d\left( \frac { x }{ y } \right) =\frac { y\quad dx-\quad x\quad dy }{ { y }^{ 2 } } ,\\ For\quad given\quad differential\quad equation,\quad divide\quad by\quad { y }^{ 2 }\quad both\quad the\quad sides\\ \therefore \quad \frac { y\quad dx-\quad x\quad dy }{ { y }^{ 2 } } +3{ x }^{ 2 }{ e }^{ { x }^{ 3 } }\quad dx=0\\ \therefore \quad d\left( \frac { x }{ y } \right) +3{ x }^{ 2 }{ e }^{ { x }^{ 3 } }\quad dx=0,\quad \\ To\quad get\quad solution\quad simply\quad integrate\quad both\quad the\quad side\quad \\ \int { d\left( \frac { x }{ y } \right) } +\quad \int { 3{ x }^{ 2 }{ e }^{ { x }^{ 3 } }\quad dx } =0\\ \frac { x }{ y } +{ e }^{ { x }^{ 3 } }=constant$

$y=mx\\ m+x\frac { dm }{ dx } =\quad m+3{ m }^{ 2 }{ x }^{ 3 }{ e }^{ { x }^{ 3 } }\\ \int { 3{ x }^{ 2 } } { e }^{ { x }^{ 3 } }dx=\quad \int { \frac { dm }{ { m }^{ 2 } } } \\ After\quad Intergrating,\\ { e }^{ { x }^{ 3 } }+C=\quad -{ m }^{ -1 }\\ \frac { x }{ y } +\quad { e }^{ { x }^{ 3 } }=\quad C$

This is a first order differential equation that can be transformed into an exact differential equation. Find an integrating factor (I used $\mu = y^{-2}$ ), integrate, then solve for the constant. Depending on how you solved, you should get $C = a\frac{x}{y} + ae^{x^3}$ where $a$ and $C$ are arbitrary constants. Since they are both constant, the equation can be rewritten as $\boxed{c = \frac{x}{y} + e^{x^3}}$ where $c=\frac{C}{a}$ is a constant.