MHB How Can You Solve This Separation of Variables Problem?

[cheatmenot]

$$(2xy-3y)dx-({x}^{2}-x)dy=0$$

ans. $$xy(x-3)=C$$

ty

 

[MarkFL]

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?

 

[cheatmenot]

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

 

[Prove It]

$$(2xy-3y)dx-({x}^{2}-x)dy=0$$

ans. $$xy(x-3)=C$$

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.

 

[MarkFL]

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?

 

[I like Serena]

$$(2xy-3y)dx-({x}^{2}-x)dy=0$$

ans. $$xy(x-3)=C$$

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.

 

[laura1231]

$\displaystyle\dfrac{2x-3}{x^2-x}dx=\dfrac{dy}{y}\Rightarrow\ \int\dfrac{2x-3}{x^2-x}dx=\int\dfrac{dy}{y}$
$y=\dfrac{kx^3}{x-1}$

 

[MarkFL]

$\displaystyle\dfrac{2x-3}{x^2-x}dx=\dfrac{dy}{y}\Rightarrow\ \int\dfrac{2x-3}{x^2-x}dx=\int\dfrac{dy}{y}$
$y=\dfrac{kx^3}{x-1}$

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.