I can find solution on fourth page here but it is not what I want. I have earlier learnt how to solve this kind of problems with serie like below. Could someone explain the integral way of doing things? Look at the border and the integrating terms, they are different -- you cannot integrate directly but use some tricks. What are they and the theory?

This is what I learnt earlier with series:

$$2\int_0^x t y(t) \; dt = x^2 + y(x)=2 \lim_{n\rightarrow 0}\sum_{k=1}^n y(\epsilon_k) \; \Delta x_k$$

where $x\in \mathbb R$, $y(x)$ is a continuous function in $\mathbb R$ to which holds $2\int_0^x ty(t) \; dt =x^2 +y(x)$ so

$$2\int_0^x t y(t) \; dt =2\lim_{n\rightarrow 0}^n \sum_{k=1}^n y\left(\frac{kx}{n}\right)\left(\frac{x}{n}\right).$$

but the solution shows an integral method. What is it (the integral method) on the page 4 called (sorry not in English, it uses simple integral apparently)?

More on page 644, X4.4, in the foreign course book I have been reading.

  • $\begingroup$ What is $f$, here? $\endgroup$ – Dylan Moreland Feb 19 '12 at 16:32
  • $\begingroup$ @DylanMoreland: it is err, I think it must be $y$ here to calculate the integral where we have an unknown variable in the border. $\endgroup$ – hhh Feb 19 '12 at 16:45

As far as I can tell, what's being done on page 4 of the link you provided, is that you are starting with the integral equation: $$ 2\int_0^ x ty(t)\,dt=x^2+y(x). $$

Then you differentiate both sides of the above with respect to $x$:
$$\tag{1} {d\over dx}\biggl[ 2\int_0^ x ty(t)\,dt\biggr]={d\over dx} \bigl(x^2+y(x)\bigr) $$

To evaluate the left-hand side derivative in the above, use the Fundamental Theorem of Calculus.

Equation $(1)$ becomes: $$ 2 xy(x) =2x+y'(x). $$ Upon rearangement, the above equation becomes: $$\tag{2} {dy\over dx}-2xy =-2x. $$

The differential equation $(2)$ is a linear first order differential equation that can be solved using an integrating factor. This is what's done on page 4 of your link (the method used is explained in the link I provided).

  • $\begingroup$ ...I see now my book misleadingly calls this on page 563 with the term "basic rule of analysis", I think it must be the same. Thanks. $\endgroup$ – hhh Feb 19 '12 at 17:52

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