Can You Solve This Non-Exact Differential Equation?

  • Thread starter freezer
  • Start date
In summary, the student is trying to find an equation that will solve for y in terms of x, but can't seem to do it. They were told to use an integrating factor, but that didn't work. They eventually found an equation that matched the original equation, but it wasn't exact.
  • #1
freezer
76
0

Homework Statement


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

Homework Equations





The Attempt at a Solution



After messing with this several times, i went through the through the textbook 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.

[itex]

\frac{\partial F}{\partial y}= y + 2e^x \neq \frac{\partial F}{\partial x} = e^x
[/itex]
 
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  • #3
I see what you are saying but not how you can simplify the origional eq into what you got.

I get

[itex]
y(\frac{y}{2}+2e^x) + (y+e^x)y' = 0
[/itex]

if you say that

p(x) = [itex] (\frac{y}{2}+2e^x) [/itex]

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

If i distribut the y'

[itex]

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

[/itex]

rearange a bit

[itex]
\frac{y^2}{2}+ yy' + 2ye^x + + y'e^x = 0
[/itex]

factor

[itex]
y(\frac{y}{2}+ y') + e^x(2y + y') = 0
[/itex]

but then what?
 
  • #4
I was thinking:

[itex]

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

[/itex]

Then:

[itex]

\frac{xy^2}{2}+ 2ye^x = -\frac{1}{2}y^2 - ye^x + c
[/itex]

then to:
[itex]
\frac{xy^2}{2}+ 2ye^x + \frac{1}{2} y^2 + ye^x = c
[/itex]
 
  • #5
freezer said:
I was thinking:

[itex]

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

[/itex]

Then:

[itex]

\frac{xy^2}{2}+ 2ye^x = -\frac{1}{2}y^2 - ye^x + c
[/itex]

You can't integrate like that since y is a function of x. You need to put it into standard form:

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

Then:

[tex]\frac{1}{N}(M_y-N_x)=1=f(x)[/tex]

The integrating factor is then [itex]e^{\int 1 dx}[/itex]

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.
 
  • #6
I did not know that Dixie Queen did this kind of calculation!
 

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