## How do you find the general solution of differential equations?

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.

 Quote by 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.
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
 $$x^{2}\frac{dy}{dx} = y^{2}$$ If you rearrange that you'd see that it equals: $$\frac{1}{y^2}dy = \frac{1}{x^2}dx$$ And not: $$y^2 dy = x^2 dx$$ To solve the equation, simply take the integral of both sides.

## How do you find the general solution of differential equations?

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)!

 Quote by 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)!
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
 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...

 Quote by 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)!
It's simple manipulation:

$$a\;\frac{b}{c}=d \Rightarrow \frac{1}{d}\;b=\frac{1}{a}\;c$$
 I thought that dy/dx is not a fraction (and therefore cannot be rearranged)?
 Recognitions: Homework Help Science Advisor 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
 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.
 Recognitions: Homework Help Science Advisor 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.
 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.
 ^ Thank you. That's what I thought!
 Recognitions: Homework Help Science Advisor 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.
 It’s not? What is it classified as then? It’s some kind of operator right?
 just a differential arn't they ? depending on whether the indeterminant is a dependent or independent variable $$dx = \Delta x$$ $$dy = f'(x) dx$$

Recognitions:
Homework Help
Hope this doesn't confuse anyone unnecessarily, but dx is the dual vector to the vector $$\partial_x$$, 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:
$$\mathcal{L}= \partial_{xy} - \partial_{x} +x^2\partial{_y}$$