Find the general solution of The ff. D.E

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Discussion Overview

The discussion revolves around finding the general solutions to two differential equations. Participants explore methods for solving these equations, including the use of partial derivatives and integrating factors. The scope includes mathematical reasoning and technical explanations related to differential equations.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents two differential equations and seeks assistance in solving them, noting that their initial attempts using partial derivatives did not yield functions of just x or just y.
  • Another participant suggests that the second equation should be a function of y alone, prompting clarification on which equation is being referenced.
  • Clarifications arise regarding the differentiation of the functions involved, with participants correcting each other's calculations of partial derivatives.
  • There is confusion expressed about the results of the differentiation, with one participant stating that neither equation simplifies to a function of just x or y.
  • Further attempts to compute the integrating factor are made, with participants discussing the expressions derived from the partial derivatives.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solvability of the equations as functions of just x or y. There are multiple viewpoints regarding the correctness of the differentiation and the resulting expressions.

Contextual Notes

Participants express uncertainty about the correctness of their differentiation steps and the implications for finding integrating factors. There are unresolved mathematical steps that may affect the overall approach to solving the differential equations.

bergausstein
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Find the general solution of The ff. D.E

1.$\displaystyle (2xy-y^2+y)dx+(3x^2-4xy+3x)dy=0$

2. $\displaystyle (x^2+y^2+1)dx+x(x-2y)dy=0$

i tried both of them using

$\displaystyle \frac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N}$

and

$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

but none of them is a function of just x or just y.

can you please help me how to go about solving this problem thanks!
 
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Look at your second option again...you should find it is a function of $y$ alone. :D
 
MarkFL said:
Look at your second option again...you should find it is a function of $y$ alone. :D

what problem are you talking about? 1 or 2?
 
bergausstein said:
what problem are you talking about? 1 or 2?

I'm sorry, I am referring to the first problem. I wanted to get that one squared away before looking at the second one. :D
 
$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

$\dfrac{\partial N}{\partial x}=-4xy$ and $\dfrac{\partial M}{\partial y}=2y$

$\frac{-4xy-2y}{2y}=y-4x$

I'm confused! help!
 
You aren't differentiating correctly:

$$\frac{\partial M}{\partial y}=2x-2y+1$$

Can you find $$\frac{\partial N}{\partial x}$$ ?
 
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$
 
bergausstein said:
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$

Correct! Now look again at the expression you want to use to compute your integrating factor...what do you find?
 

$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

$\frac{6x-4x+3-(2x-2y+1)}{2xy-y^2+y}=\frac{2y+2}{2xy-y^2+1}$ there's still x here!
 
Last edited:
  • #10
Try the second option...
 
  • #11
both options are not a function of just x and y.

as I stated in my OP. :(
 
  • #12
Let's take a look...

$$\frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}=\frac{(6x-4y+3)-(2x-2y+1)}{2xy-y^2+y}=\frac{4x-2y+2}{y(2x-y+1)}$$

Now factor the numerator...:D
 

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