Question

can anyone outline why dy/dx -3y/x=7 is the derivative of y=-7x/2 + kx^3?

Answer 1

Is this your equation?\[y = \frac{-7x}{2} + kx^3\]

Answer 2

@charlymix ?

Answer 3

yep.. sorry

Answer 4

Well then\[\frac{dy}{dx} = \frac{-7}{2} + 3kx^2\] [Done with good ol' differentiation :D ]

Now you have to verify,\[\frac{dy}{dx} = 7 + \frac{-3y}{x}\]

I think you ought to substitute y's original value and check if it is correct! :)

Answer 5

ok, actually what I don't understand is why the factor e ^∫Pdx lead us to the primitive, and the original value is unknown. If you can give some example???

Answer 6

\[\frac{dy}{dx}-\frac{3}{x}y=7\]
The integrating factor is \(\large e^{-\int 3/x~dx}=\dfrac{1}{x^3}\), and so
\[\frac{1}{x^3}\frac{dy}{dx}-\frac{3}{x^4}y=\frac{7}{x^3}\\
\frac{d}{dx}\left[\frac{1}{x^3}y\right]=\frac{7}{x^3}\\
\frac{1}{x^3}y=\int\frac{7}{x^3}~dx\\
\frac{1}{x^3}y=-\frac{7}{2x^2}+k\\
y=-\frac{7}{x}+kx^3\]
To address your more recent question, consider a general first order linear ODE,
\[f(x)\frac{dy}{dx}+g(x)y=h(x)\]
Let's isolate the derivative function and rename the functions:
\[\frac{dy}{dx}+P(x)y=Q(x)\]
The integrating factor is \(\large e^{\int P(x)~dx}\), so you have
\[e^{\int P(x)~dx}\frac{dy}{dx}+e^{\int P(x)~dx}P(x)y=e^{\int P(x)~dx}Q(x)\]
Notice that the right hand side is the result of taking a derivative, namely of the product of the integrating factor and \(y\):
\[\begin{align*}\frac{d}{dx}\left[e^{\int P(x)~dx}y\right]&=e^{\int P(x)~dx}\frac{dy}{dx}+y\frac{d}{dx}\left[e^{\int P(x)~dx}\right]\\
&=e^{\int P(x)~dx}\frac{dy}{dx}+y~e^{\int P(x)~dx}\frac{d}{dx}\left[\int P(x)~dx\right]\\
&=e^{\int P(x)~dx}\frac{dy}{dx}+e^{\int P(x)~dx}P(x)y
\end{align*}\]

Answer 7

thank you... I'm going to study the whole process because I like to know the underlying of stuffs. One more question, where does the integrating factor come from?

Answer 8

The integrating factor is more something we conjure rather than derive from the given equation. Like many of the methods used for solving DE's, the use of the factor probably arose as a result of noticing a pattern; in this case, it's the pattern of differentiating a product.

If you look at the last equation I have up there, you'll see that it's exactly the right hand side multiplied by the integrating factor. Without the integrating factor, you can't do much in the way of proceeding with finding a solution.

Answer 9

All right, that's what I was imagining :) I'd rather concentrate on the last equation... thanks again!

Answer 10

yw