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

Zack Chase
2022-09-22
Answered

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

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Matthias Calhoun

Answered 2022-09-23
Author has **11** answers

Find the general solution of

$xdy+(3xy+y-{e}^{-3x})dx=0$

we have a differential equation as

$(3xy+y-{e}^{-3x})dx+xdy=0$ (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 ${e}^{3x}\text{to}E{q}^{n}$ (i)

$(3xy{e}^{3x}+{e}^{3x}y-1)dx+x\cdot {e}^{3x}\cdot dy=0$

Solution: $\int (3xy{e}^{3x}+{e}^{3x}y-1)dx+\int N\cdot dy=C\phantom{\rule{0ex}{0ex}}\Rightarrow y\cdot \int 3x\cdot {e}^{3x}\cdot dx+y\frac{{e}^{3x}}{3}-x+0=c\phantom{\rule{0ex}{0ex}}\Rightarrow y\cdot x\cdot {e}^{3x}-\int 3\frac{{e}^{3x}}{3}\cdot dx+y\cdot \frac{{e}^{3x}}{3}-x=c\phantom{\rule{0ex}{0ex}}\Rightarrow xy{e}^{3x}-\frac{{e}^{3x}\cdot y}{3}+\frac{{e}^{3x}\cdot y}{3}-x=c\phantom{\rule{0ex}{0ex}}\Rightarrow xy{e}^{3x}-x=c$

$xdy+(3xy+y-{e}^{-3x})dx=0$

we have a differential equation as

$(3xy+y-{e}^{-3x})dx+xdy=0$ (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 ${e}^{3x}\text{to}E{q}^{n}$ (i)

$(3xy{e}^{3x}+{e}^{3x}y-1)dx+x\cdot {e}^{3x}\cdot dy=0$

Solution: $\int (3xy{e}^{3x}+{e}^{3x}y-1)dx+\int N\cdot dy=C\phantom{\rule{0ex}{0ex}}\Rightarrow y\cdot \int 3x\cdot {e}^{3x}\cdot dx+y\frac{{e}^{3x}}{3}-x+0=c\phantom{\rule{0ex}{0ex}}\Rightarrow y\cdot x\cdot {e}^{3x}-\int 3\frac{{e}^{3x}}{3}\cdot dx+y\cdot \frac{{e}^{3x}}{3}-x=c\phantom{\rule{0ex}{0ex}}\Rightarrow xy{e}^{3x}-\frac{{e}^{3x}\cdot y}{3}+\frac{{e}^{3x}\cdot y}{3}-x=c\phantom{\rule{0ex}{0ex}}\Rightarrow xy{e}^{3x}-x=c$

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