Question

ydx-xdy+3x^2y^2e^x^3 dx=0​

Answer 1

Given Question :-

Solve the differential equation

$\rm :\longmapsto\:ydx - xdy + {3x}^{2} {y}^{2} {( {e}^{ {x}^{3} })}dx = 0$

Formula Used :-

$\boxed{ \red{ \bf \:\dfrac{d}{dx}\dfrac{x}{y} = \dfrac{ydx - xdy}{ {y}^{2} }}}$

Solution :-

$\rm :\longmapsto\:ydx - xdy + {3x}^{2} {y}^{2} {( {e}^{ {x}^{3} })}dx = 0$

$\rm :\longmapsto\:(ydx - xdy) = - {3x}^{2} {y}^{2} {( {e}^{ {x}^{3} })}dx$

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \red{ \bf \:Divide \: by \: {y}^{2}, \: we \: get}}$

$\rm :\longmapsto\:\dfrac{ydx - xdy}{ {y}^{2} } = - {3x}^{2}{( {e}^{ {x}^{3} })}dx$

$\rm :\longmapsto\:d\bigg( \dfrac{x}{y} \bigg) = - d \: ( {e}^{ {x}^{3} })$

$\rm :\longmapsto\:\bigg( \dfrac{x}{y} \bigg) = - \: ( {e}^{ {x}^{3} }) + c$

$\purple{\bf\implies \:\:\bigg( \dfrac{x}{y} \bigg) + \: ( {e}^{ {x}^{3} }) = c}$

Additional Information :-

The differential equation of the form Mdx + Ndy = 0 is exact Differential equation if

$\boxed{ \red{ \bf \:\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}}}$

and solution is given by

$\boxed{ \red{ \bf \:\int_{y \: constant} Mdx+ \int \:N (term \: not \: containing \: x)dy}} = c$