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differentiating $f(x,y)$, w.r.t y, we have, $f_y^{'}(x,y)=2y$

differentiating $g(y)$, w.r.t x, we have, $g_x^{'}(y)=0$

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You do realize the equation is not exact (as is), right? – JohnD Feb 15 '13 at 3:37
i was looking for the integrating factor by visual inspection. – Rajesh K Singh Feb 15 '13 at 3:43
Ok, I just don't know what your thoughts are unless you tell us. – JohnD Feb 15 '13 at 3:45
up vote 2 down vote accepted

We have the differential equation:

$$(x^2 + y^2 +2x) + (2y)y' = 0 \tag{1}$$

If we let: $$f(x, y) = x^2 + y^2 + 2x \\ g(x,y) = 2y$$

Then: $$f_y = 2y\\g_x = 0$$

"Oh No! It's not exact, as we hoped!" However, there is still a chance to save the equation. Let's try the integrating factor method.

If we let: $$\frac{d\mu}{dx} = \frac{f_y - g_x}{g}\mu = \frac{2y}{2y}\mu$$ $$\frac{d\mu}{dx} = \mu$$

Solving, $\mu = e^x$.

Multiply (1) by the integrating factor $\mu$: $$e^x(x^2 + y^2 +2x) + (2ye^x)y' = 0 \tag{2}$$

Now we have two new functions, let's call them $M$ and $N$: $$M(x, y) = e^x(x^2 + y^2 +2x)\\N(x, y) = (2ye^x)$$ $$M_y = 2ye^x = N_x$$ Therefore, equation (2) is exact. Solve using the typical method. (If you need guidance past this point, please let me know via comment.)

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thanks for the prompt and correct respose !! – Rajesh K Singh Feb 15 '13 at 3:39

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