# DQ worksheet equation help

## Homework Statement

This is from a worksheet...
$(\frac{y^2}{2}+ 2ye^x) + (y + e^x)\frac{_{dy}}{dx} = 0$

## The Attempt at a Solution

After messing with this several times, i went through the through the text book for examples but nothing in the chapter matches with the exception of exact.. with a bit of manipulation i matched up the form. But it is not exact so that does not work. We have covered up separable, substitution, and exact and none seem to fit.

$\frac{\partial F}{\partial y}= y + 2e^x \neq \frac{\partial F}{\partial x} = e^x$

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rock.freak667
Homework Helper
I see what you are saying but not how you can simplify the origional eq into what you got.

I get

$y(\frac{y}{2}+2e^x) + (y+e^x)y' = 0$

if you say that

p(x) = $(\frac{y}{2}+2e^x)$

then p(x) is not going to integrate very nice.

If i distribut the y'

$\frac{y^2}{2}+2ye^x + yy'+ y'e^x = 0$

rearange a bit

$\frac{y^2}{2}+ yy' + 2ye^x + + y'e^x = 0$

factor

$y(\frac{y}{2}+ y') + e^x(2y + y') = 0$

but then what?

I was thinking:

$(\frac{y^2}{2}+ 2ye^x)dx = -(y+e^x)dy$

Then:

$\frac{xy^2}{2}+ 2ye^x = -\frac{1}{2}y^2 - ye^x + c$

then to:
$\frac{xy^2}{2}+ 2ye^x + \frac{1}{2} y^2 + ye^x = c$

I was thinking:

$(\frac{y^2}{2}+ 2ye^x)dx = -(y+e^x)dy$

Then:

$\frac{xy^2}{2}+ 2ye^x = -\frac{1}{2}y^2 - ye^x + c$
You can't integrate like that since y is a function of x. You need to put it into standard form:

$$(y^2/2+2ye^x)dx+(y+e^x) dy=0=Mdx+Ndy$$

Then:

$$\frac{1}{N}(M_y-N_x)=1=f(x)$$

The integrating factor is then $e^{\int 1 dx}$

Now multiply both sides of the standard-form by this integrating factor to make it exact then proceed to solve it using the technique of exact equations. Also, first check that it is indeed exact in case I made a mistake.

HallsofIvy