Find the general solution of The ff. D.E

In summary, the general solution of the given differential equations is obtained by using the formula $\frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$ and factoring the numerator of the second equation.
  • #1
bergausstein
191
0
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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  • #2
Look at your second option again...you should find it is a function of $y$ alone. :D
 
  • #3
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?
 
  • #4
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
 
  • #5
$\displaystyle \frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}$

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

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

I'm confused! help!
 
  • #6
You aren't differentiating correctly:

\(\displaystyle \frac{\partial M}{\partial y}=2x-2y+1\)

Can you find \(\displaystyle \frac{\partial N}{\partial x}\) ?
 
  • #7
$\displaystyle \frac{\partial N}{\partial x}=6x-4x+3$
 
  • #8
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?
 
  • #9

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

$\frac{6x-4x+3-(2x-2y+1)}{2xy-y^2+y}=\frac{2y+2}{2xy-y^2+1}$ 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...

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

Now factor the numerator...:D
 

FAQ: Find the general solution of The ff. D.E

1. What is a general solution in differential equations?

A general solution is a mathematical expression that satisfies a given differential equation and includes all possible solutions to the equation. It typically contains arbitrary constants that can take on any value.

2. How do you find the general solution of a differential equation?

To find the general solution of a differential equation, you need to first solve the equation by finding the antiderivative of the function. This will result in a general solution that includes arbitrary constants. You can then use initial conditions or boundary conditions to find the specific solution that satisfies the given conditions.

3. What is the difference between a general solution and a particular solution?

A general solution includes all possible solutions to a differential equation, while a particular solution is a specific solution that satisfies given initial conditions or boundary conditions. A particular solution can be obtained from the general solution by substituting in the appropriate values for the arbitrary constants.

4. Can a differential equation have more than one general solution?

No, a differential equation can only have one general solution. However, this solution may include arbitrary constants that can take on different values, resulting in an infinite number of particular solutions.

5. What is the importance of finding the general solution of a differential equation?

Finding the general solution allows us to understand the behavior of a system over time. It also helps us to find the particular solution that satisfies given initial conditions or boundary conditions, which can then be used to make predictions and solve real-world problems.

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