Basic Differential Equations

[Jason76]

Aren't they simply indefinite integration at the basic level? They seem to be the same thing.

Ex.

Find the general solution;

\(\displaystyle \dfrac{dy}{dx} = x^{2} + 1\)

\(\displaystyle \int dy = \int x^{2} + 1 dx\)

\(\displaystyle y = \dfrac{{x}^{3}}{3} + C\)

You do indefinite integration to get the general solution, but the general solution is simply a solved indefinite integration problem. Now, finding a particular solution is something found only in differential equations courses.

 

[tkhunny]

At this level, yes. It's just a little different look at it. No worries. There will be lots more to learn.

 

[soroban]

Hello, Jason76!

Careful!

\(\displaystyle \dfrac{dy}{dx} = x^{2} + 1\)

\(\displaystyle \displaystyle\int dy \:=\: \int \left(x^2 + 1\right)\,dx\)

. . \(\displaystyle y \:=\: \dfrac{x^3}{3} \color{red}{+ x} + C\)

 

[Jason76]

Hello, Jason76!

Careful!




\(\displaystyle \displaystyle\int dy \:=\: \int \left(x^2 + 1\right)\,dx\)


. . \(\displaystyle y \:=\: \dfrac{x^3}{3} \color{red}{+ x} + C\)

Ok, got it. Constants with no number connected become the constant with x (constant and x are connected) when integrated. For instance, the integral of 5 is 5x, and the integral of 1 is 1x or x.

 

[HallsofIvy]

Yes, at that very basic level, \(\displaystyle \frac{dy}{dx}= f(x)\) you are saying "the derivative of y is f(x)" so finding y is just finding the "anti-derivative". But, very quickly, you need to start looking at things like \(\displaystyle \frac{dy}{dx}= f(x, y)\).