Variable Separable Method Of Differential Equations
Whenever it comes to solving Differential Equations, Initially almost all of us try to solve it by separating the variables as it is the most easy and convenient way for the majority of the students. So, here we'll try to understand this Variable separable method in a more clear way.
Do watch this video by Jeff Chasnov sir for making this concept more clear:
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A first-order ode is separable if it can be written in the form:
where the function g(y) is independent of x and f (x) is independent of y. Integration from xo to x
The integral on the left can be transformed by substituting u = y(x), du = y'(x)dx, and changing the
lower and upper limits of integration to y(x0) = y0 and y(x) = y. Therefore,
which can often yield an analytical expression for y = y(x) if the integrals can be done and the
resulting algebraic equation can be solved for y.
A simpler procedure that yields the same result is to treat dy/dx as a fraction. Multiplying the
differential equation by dx results directly in,
g(y) dy = f (x) dx
So, let's try and learn how to solve some first-order differential equations that occur actually quite often in applications. It's a first order equation. The equation looks like dy dx, and then we're going to have a right-hand side then that splits into two terms. The term on the right-hand side I will say is f of x, and then on the left-hand side, we'll have a g of y. Basically, the right-hand side is f of x divided by g of y, or you can say that it's just some function of x times some function of y. Together with this we need an initial condition. Let's say y of x naught equals y naught. If first-order equation is of this form where the right-hand side splits into a function of x times a function of y, and we can write it like this, then we say that the equation is separable. A separable equation can be integrated. We integrate this equation. We integrate from the initial value of x. So, we integrate from x naught to x, and then we have g of y which is a function of x times dy dx, dy dx is y prime of x, and then we're integrating this dx. That's the integral of the left-hand side with respect to x is equal to the integral of the right-hand side which is x naught to x f of x dx. The equation is separable and we integrate it. You can notice that the left-hand side of this equation looks like a typical substitution type of integration. We can let u equals y of x, that's our substitution variable, and then du becomes y prime of x dx. Using this substitution we write the integral. When x equals x naught, then we have y of x naught, which will be y naught, that's our initial condition, y of x naught is y naught. We integrate from x naught to x, y of x is what we call y, and then we have g of u du with the substitutions, and that's equal to the integral from x naught to x f of x dx. The first-order equation that's separable written in the form g of y dy dx equals f of x, we can simply integrate and end up with a solution that means that you have an integral to do on the left and integral to do on the right, and then in many cases, you can then solve this equation for y as a function of x. That's the separable first-order equation. In practice, one doesn't do this substitution. In practice, it's much more straightforward to solve an equation like this. Let me show you what one does in practice. You have this equation, g of y dy dx equals f of x. What you do is you'd may take advantage of this Leibniz's notation for the derivative. We know dy dx is not really a fraction, it just means the derivative of y with respect to x. But in this particular situation, you can treat dy dx like a fraction. That will eliminate this substitution step. If you treat dy dx like a fraction, you can write this as g of y dy equals f of x dx. That's the separation stage where you have a function of y times dy equals a function of x times dx. It's very important here that you have this dy is multiplying everything here on the left, and there's no x dependence on the left, and this dx is multiplying everything here on the right, and there is no y dependence on the right. Getting to this stage, you then integrate. Here we integrate from x naught to x, that will be our right-hand side here. On the left, when you integrate from x naught to x using the initial condition here when x equals x naught, y is equal to y naught, so you integrate from y naught, and when x is the value of x then y is the value of y, we integrate to y. This equation is the same as this one below. Here u is a dummy variable, so with commonly just use y as the dummy variable. Y inside the integral has no meaning other than holding variable, and the true y here is in the limit. That's the very very quick solution of separable equation. The only thing that's important here is that you have to manipulate the original differential equation. So you have some function of y times dy dx, equals some function of x, and then you can multiply through by dx. Let me quickly review. This is a class of first-order equations where it's possible to find an analytical solution, is called separable first-order differential equations. The type of equation is such that this general right-hand side which can be a function of x and y is possible to separate it into just a function of x times a function of y, and then we can pull the y dependence on one side and the x dependence on the other, and then treat dy dx as if it was a fraction and then integrate. We'll see some examples of how to solve equations of this type. I'm Jeff Chasnov, thanks for watching, and I'll see you in the next video.
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