How Can You Solve This Separation of Variables Problem?

I would like to see the OP work through the problem, and we can provide guidance as needed. I don't believe it is helpful to just provide a solution without the OP having worked the problem at least to some extent.
  • #1
cheatmenot
4
0
\(\displaystyle (2xy-3y)dx-({x}^{2}-x)dy=0\)

ans. \(\displaystyle xy(x-3)=C\)

ty
 
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  • #2
Hello and welcome to MHB! :D

I have shortened the thread title a bit and moved it here to our Differential Equations forum.

What do you get when you separate variables?
 
  • #3
i need the solution in the given problem . .i found out in my book the answer . .i need the solution on how to solve the problem using the separation of variables
 
  • #4
cheatmenot said:
\(\displaystyle (2xy-3y)dx-({x}^{2}-x)dy=0\)

ans. \(\displaystyle xy(x-3)=C\)

ty

$\displaystyle \begin{align*} \left( 2\,x\,y - 3\,y \right) \, \mathrm{d}x - \left( x^2 - x \right) \, \mathrm{d}y &= 0 \\ \left( 2\, x\, y -3\,y \right) \, \mathrm{d}x &= \left( x^2 - x \right) \, \mathrm{d}y \\ 2\,x\,y - 3\,y &= \left( x^2 - x \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} \\ y \, \left( 2\,x - 3 \right) &= \left( x^2 - x \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{2x - 3}{x^2 - x} &= \frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{2x - 3}{x \, \left( x - 1 \right) } &= \frac{1}{y} \, \frac{\mathrm{d}y}{\mathrm{d}x} \\ \int{ \frac{2x - 3}{x \, \left( x - 1 \right) }\, \mathrm{d}x } &= \int{ \frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x} \\ \int{ \frac{2x - 3}{ x \, \left( x - 1 \right) } \, \mathrm{d}x} &= \int{ \frac{1}{y}\,\mathrm{d}y} \end{align*}$

Go from here. You will need to use Partial Fractions on the left hand side.
 
  • #5
cheatmenot said:
i need the solution in the given problem . .i found out in my book the answer . .i need the solution on how to solve the problem using the separation of variables

Our goal here is to help students work their problems so that they are an active participant in the process of obtaining a solution and learn more that way, not just work the problem for them.

So, can you state what you get after you separate the variables?
 
  • #6
cheatmenot said:
\(\displaystyle (2xy-3y)dx-({x}^{2}-x)dy=0\)

ans. \(\displaystyle xy(x-3)=C\)

ty

You seem to have a typo in either your problem statement or your answer.

Suppose we verify the answer, then we get:
$$d(xy(x-3)) = dx\,y(x-3) + x\, dy\,(x-3) + xy\,dx = (2xy-3y)dx + (x^2-3x)dy = 0$$

As you can see, this does not match your problem statement.
 
  • #7
$\displaystyle\dfrac{2x-3}{x^2-x}dx=\dfrac{dy}{y}\Rightarrow\ \int\dfrac{2x-3}{x^2-x}dx=\int\dfrac{dy}{y}$
$y=\dfrac{kx^3}{x-1}$
 
  • #8
laura123 said:
$\displaystyle\dfrac{2x-3}{x^2-x}dx=\dfrac{dy}{y}\Rightarrow\ \int\dfrac{2x-3}{x^2-x}dx=\int\dfrac{dy}{y}$
$y=\dfrac{kx^3}{x-1}$

There are most likely many people here who can work this problem, but it would be better for the OP to get the OP to work it on their own, with some guidance. If we just give the solution, we have provided little service.
 

FAQ: How Can You Solve This Separation of Variables Problem?

What is the "Separation of Variables Problem"?

The separation of variables problem is a mathematical technique used to solve differential equations. It involves breaking down a complex equation into simpler equations that can be solved separately, and then combining those solutions to find the overall solution.

Why is "Separation of Variables" important in science?

Separation of variables is important in science because it allows us to find solutions to complex problems that would otherwise be difficult or impossible to solve. It is particularly useful in fields such as physics and engineering where differential equations are commonly used to model real-world systems.

What are some common applications of "Separation of Variables"?

Some common applications of separation of variables include solving heat conduction equations, wave equations, and Schrödinger's equation in quantum mechanics. It is also used in fields such as fluid dynamics, electromagnetism, and astrophysics.

What are the steps involved in solving a "Separation of Variables" problem?

The first step is to identify the differential equation and determine if it can be solved using separation of variables. Then, the equation is separated into simpler equations by isolating the variables on opposite sides of the equation. Next, each equation is solved separately using integration or other mathematical techniques. Finally, the solutions are combined to find the overall solution to the original equation.

What are the limitations of "Separation of Variables"?

While separation of variables is a powerful technique, it is not always applicable to all types of differential equations. It can only be used for linear, homogeneous equations, and the variables must be separable. Additionally, the technique may not always yield a complete solution and may require additional conditions or assumptions to be made.

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