Find the general solution of xdy + (3xy + y - e^(-3x)) * dx = 0

Find the general solution of $xdy+\left(3xy+y-{e}^{-3x}\right)dx=0$
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Matthias Calhoun
Find the general solution of
$xdy+\left(3xy+y-{e}^{-3x}\right)dx=0$
we have a differential equation as
$\left(3xy+y-{e}^{-3x}\right)dx+xdy=0$ (i)
$\frac{\mathrm{\partial }M}{\mathrm{\partial }y}=3x+1,\frac{\mathrm{\partial }N}{\mathrm{\partial }x}=1$
Now, $\frac{\frac{\mathrm{\partial }M}{\mathrm{\partial }y}-\frac{\mathrm{\partial }N}{\mathrm{\partial }x}}{N}=3$
$I\cdot f={e}^{\int 3\cdot dx}={e}^{3x}$
Multiply by (i)
$\left(3xy{e}^{3x}+{e}^{3x}y-1\right)dx+x\cdot {e}^{3x}\cdot dy=0$
Solution: $\int \left(3xy{e}^{3x}+{e}^{3x}y-1\right)dx+\int N\cdot dy=C\phantom{\rule{0ex}{0ex}}⇒y\cdot \int 3x\cdot {e}^{3x}\cdot dx+y\frac{{e}^{3x}}{3}-x+0=c\phantom{\rule{0ex}{0ex}}⇒y\cdot x\cdot {e}^{3x}-\int 3\frac{{e}^{3x}}{3}\cdot dx+y\cdot \frac{{e}^{3x}}{3}-x=c\phantom{\rule{0ex}{0ex}}⇒xy{e}^{3x}-\frac{{e}^{3x}\cdot y}{3}+\frac{{e}^{3x}\cdot y}{3}-x=c\phantom{\rule{0ex}{0ex}}⇒xy{e}^{3x}-x=c$