Picard theorem

[faik mermer]

hello guys The question will be solved with the picard method, but I could not create a yn (x) formula.

 

[faik mermer]

hello guys The question will be solved with the picard method, but I could not create a yn (x) formula.

 

[HallsofIvy]

The differential equation is dy/dx= f(x,y)= xy with the initial condition, y(0)= 3.
We can write the differential equation as dy= f(x,y)dx= xydx and the integrate both sides:
$\displaystyle y= \int_0^x f(t,y)dt+ 3$ except that, of course we don't know y on the right side to integrate it!

Picard's method starts by, first, replacing y by its initial condition. Here we take $\displaystyle y_0= 3$ and then calculate the next y as $\displaystyle y_1= \int_0^x 3tdt+ 3= \frac{3}{2}x^2+ 3$. Then the next y is $\displaystyle y_2= \int_0^x t\left(\frac{3}{2}t^2+ 3\right)dt+ 3= \int_0^x \frac{3}{2}t^3+ 3t dt+ 3= \frac{3}{8}x^4+ \frac{3}{2}x^2+ 3$. $\displaystyle y_3= \int_0^x t\left(\frac{3}{8}t^4+ \frac{3}{2}t^2+ 3\right)dt+ 3= \int_0^x\left(\frac{3}{8}t^5(+ \frac{3}{2}t^3+ 3t\right)dt+ 3= \frac{3}{48}x^6+ \frac{3}{8}t^4+ \frac{3}{2}t^2+ 3$.

Is that enough to "guess" The general form? The powers of x are all even. The numerators of the fractions are "3". The denominators are the hard part. They are 2= 1(2), 8= (2)(4), and 48= (3)(16). I might "guess" that the next denominator is 4(32)= 128 so that the next term is $\displaystyle \frac{3}{128}x^8$. Check if that is correct. So what would the general form be?