faik mermer
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Picard theorem
- Thread starter faik mermer
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faik mermer
New member
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HallsofIvy
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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?
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?