-b.2.2.1 separate variables y'=\dfrac{x^2}{y}

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In summary, the purpose of separating variables in this equation is to make it easier to solve for the dependent variable, y. This can be done by moving all terms involving y to one side and all terms involving x to the other side. The general solution to this differential equation is y = Cx^2, where C is a constant of integration. This equation can be solved using the separation of variables method, but there are limitations such as only being applicable for first-order equations in a specific form and not working for more complex equations.
  • #1
karush
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MHB
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$\tiny{b.2.2.1 \quad 48}$
320
solve $\quad y'=\dfrac{x^2}{y}$

$\begin{array}{lll}
\textit{Rewrite as}
&y\dfrac{dy}{dx}=x^2\implies y\ dy=x^2\ dx \\ \\
\textit{Integrate Thru}
&\displaystyle\int y \, dy = \displaystyle\int x^2 \, dx\\ \\
&\dfrac{y^2}{2} = \dfrac{x^3}{3}+c\\ \\
\textit{thus}
&3y^2-2x^3+c; \quad y\ne 0
\end{array}$

I think this is OK but always get confused about the c
 
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  • #2
Well done.

Two things...
1) \(\displaystyle 3y^2 - 2x^3 + c\) is not an equation.

2) \(\displaystyle \dfrac{y^2}{2} = \dfrac{x^3}{3} + c\) and \(\displaystyle 3y^2 - 2x^3 + c = 0\) have two different values for c. You can keep it like this and use c for both (almost everyone does) but you really should mention that you are redefining the value of c.

-Dan
 
  • #3
ok looks like left out the =$3y^2-2x^3=c; \quad y\ne 0$
this was the book answer :cool:
 
  • #4
I made a pdf of 22 differential equation problems with replies from MHB
it is still in draft mode with markup corrections to be made but so far these have attracted 7K+ views:rolleyes:

hope it opens!

https://drive.google.com/file/d/1ARMDHlIAnGE6DEpUfDL6D4lNndt0OFeq/view?usp=sharing
 
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  • #5
Yes, integration of $\int y dy= \int x^2dx$ gives $\frac{y^2}{2}= \frac{x^3}{3}+ c$
where "c" is the "constant of integration" and can be any number.

Now clear the fractions by multiplying by 6:
$3y^2= 2x^3+ 6c$.

Since c can be any constant, so can 6c. Sometimes people just write
$3y^2= 2x^3+ c$ again, understanding that this "c" is not the same number as the first "c" but is still just an undetermined constant. Some people prefer to write "c'" or "C" to make it clear that this is not the same number.

We could just as well have written the original integration as $\frac{y^2}{2}= \frac{x^3}{3}+ \frac{c}{6}$ since "c/6" is just as much an undermined constant as "c" is! Then multiplying by 6 give $3y^2= 2x^3+ c$.
 
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Related to -b.2.2.1 separate variables y'=\dfrac{x^2}{y}

1. What is the purpose of separating variables in a differential equation?

Separating variables allows us to solve a differential equation by expressing the dependent and independent variables on opposite sides of the equation. This allows us to integrate both sides separately, making it easier to find a solution.

2. How do you separate variables in a differential equation?

To separate variables, we need to rearrange the equation so that all terms containing the dependent variable are on one side and all terms containing the independent variable are on the other side. This usually involves multiplying or dividing both sides by a certain factor.

3. Can you explain the steps for solving a differential equation using the separate variables method?

First, we need to separate the variables as described in the previous answer. Then, we integrate both sides of the equation with respect to their respective variables. This will give us an equation with an arbitrary constant. We can then use initial conditions or other information to solve for the constant and find the particular solution.

4. Are there any limitations to using the separate variables method for solving differential equations?

The separate variables method can only be used for certain types of differential equations, specifically those that can be written in the form y' = f(x)g(y). It also may not work for more complex or nonlinear equations.

5. Can you provide an example of solving a differential equation using the separate variables method?

Sure, let's take the equation y' = x/y as an example. We can rearrange it as ydy = xdx, then integrate both sides to get (y^2)/2 = (x^2)/2 + C. Using the initial condition y(0) = 1, we can solve for C and get the particular solution y = √(x^2 + 2).

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