How do you find the general solution of differential equations?

Fritz

I really don't understand the concept of differential equations! I'm confused about how you would go about solving them.

if (x^2).dy/dx = y^2,

integral(y^2)dy = integral(x^2)

1/3y^3 = 1/3x^3 + A

^ What have I done wrong here?


I don't get the whole thing where you separate the variables etc.

Am I right in saying that integral(dy/dx)dx = y?



Moderators, sorry I posted this twice. It was an accident.
Please can someone help me understand all of this? I understand basic differentiation and integration.

marlon

Just put all variables of one sort at one side : x²dy/dx = y². This means that (1/y²)dy = (1/x²)dx and just integrate --> -(1/y) + C = -(1/x) + C'

marlon

devious_

[tex]x^{2}\frac{dy}{dx} = y^{2}[/tex]
If you rearrange that you'd see that it equals:
[tex]\frac{1}{y^2}dy = \frac{1}{x^2}dx[/tex]
And not:
[tex]y^2 dy = x^2 dx[/tex]

To solve the equation, simply take the integral of both sides.

Fritz

So you substitute y for x and vice versa, take the reciprocal of both variables and then integrate?

I don't understand how you got from the first step to the second step (and then to the third)!

marlon

well the integral of (1/x²) is -(1/x), that's all...


use the rule integral of x^n = (x^(n+1)/n+1) and n = -2 here...
marlon

Fritz

I understand that, but I don't get why you substituted x^2 for 1/y^2 and y^2 for 1/x^2.

It's not integrating of individual terms I'm confused about...

devious_

It's simple manipulation:

[tex]a\;\frac{b}{c}=d \Rightarrow \frac{1}{d}\;b=\frac{1}{a}\;c[/tex]

Fritz

I thought that dy/dx is not a fraction (and therefore cannot be rearranged)?

matt grime

dy/dx is indeed not a fraction, but there are *some* circumstances where it can be treated as one, and this is one of them. We're just simply taking a practical approach to how to solve the equation, that is omitting some steps that always occur in the same way, to clear up the presentation.

What it is is a short hand wayo of saying that if:

f(y)dy/dx = g(x)

after rearranging, then

int f dy = int g dx

Fritz

What do f dy and g dx mean? Don't you mean int f(y)dy = int f(x)dx?

I don't get what f and g on their own mean.

matt grime

yes f on its own means f(y), but on the right hand side it is int g(x)dx

normally, where there is no room for confusion, it is acceptable, for example, to drop the letter t from h(t) and refer to the function simply as h.

JonF

You can treat the dy/dx like a fraction because it essentially is one. It is the ratio of two differential operators. So when you multiply both sides by the differential operator dx, you get cancellation on one side and a dx on the other.

Fritz

^ Thank you. That's what I thought!

matt grime

dx isn't a differential operator (it doesn't take a function and yield another function, but you don't want to get too technical about that) they are referred to as infinitesimals occasionally.

JonF

It’s not? What is it classified as then? It’s some kind of operator right?

gazzo

just a differential arn't they ?

depending on whether the indeterminant is a dependent or independent variable

[tex]dx = \Delta x[/tex]
[tex]dy = f'(x) dx[/tex]

matt grime

Hope this doesn't confuse anyone unnecessarily, but dx is the dual vector to the vector [tex]\partial_x[/tex], and is called a 1-form. It is an element of the cotangent bundle, but it is not a differential operator. (Most things can be thought of as "operators" I suppose, it's one of those frequently used ambiguous labels). A differential operator looks something like:

[tex] \mathcal{L}= \partial_{xy} - \partial_{x} +x^2\partial{_y}[/tex]