Solve

$\displaystyle (x^2 + y^2 + x)dx=-xydy$

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Malay

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- Sep 13th 2007, 06:18 PMmalaygoel[SOLVED] differential equation
Solve

$\displaystyle (x^2 + y^2 + x)dx=-xydy$

Keep Smiling

Malay - Sep 13th 2007, 06:33 PMJhevon
recall that exact differential equations are those of the form $\displaystyle M(x,y)~dx + N(x,y)~dy = 0$, where $\displaystyle \frac {\partial M}{\partial y} = M_y = N_x = \frac {\partial N}{\partial x}$

So this guy starts off looking like an exact equation, right, with $\displaystyle M(x,y) = x^2 + y^2 + x$ and $\displaystyle N(x,y) = xy$

but lo and behold, $\displaystyle M_y = 2y \ne N_x = y$. thus we need to find an integrating factor

Recall that we can find the integrating factor $\displaystyle \mu$ by the differential equation:

$\displaystyle \frac {d \mu}{dx} = \frac {M_y - N_x}{N} \mu$

so here we have:

$\displaystyle \frac {d \mu}{dx} = \frac {2y - y}{xy} \mu = \frac {\mu}{x}$

$\displaystyle \Rightarrow \mu (x) = x$

multiplying through by $\displaystyle x$ we obtain:

$\displaystyle \left( x^3 + xy^2 + x^2 \right)~dx + \left( x^2y \right)~dy = 0$

which is now an exact equation, since $\displaystyle M_y = 2xy = N_x$

can you continue?