Suvadip
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if
=(1+x)^2,%20y_2(x)=\frac{2}{3}x^3+3x^2+2x+1,%20y_3(x)=\frac{1}{3}x^4+2x^3+3x^2+2x+1)
, then what will be
)
. In fact I was solving the integral equation
=1+2\int_0^x(t+y(t))dt,%20y_0(x)=1)
by the method of successive approximation.
Integral equation by successive approximation
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Siron
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I was not familiar with the name of the method until I noticed it's the same as Picard Iterations (after some googling). I tried some ways to see a pattern in the polynomials but I failed. Especially the leading terms are very unpredictable in my opinion. Perhaps it's not possible to find an explicit form for $y_n(x)$.
What's the background of this problem?
What's the background of this problem?
chisigma
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suvadip said:if, then what will be. In fact I was solving the integral equationby the method of successive approximation.
Under appropriate conditions the solution of the differential equation ...
$\displaystyle y^{\ '} = f(x,y),\ y(0)= y_{0}\ (1)$
... must satisfy the following integral equation...
$\displaystyle y = y_{0} + \int_{x_{0}}^{x} f\{t, y(t)\}\ d t\ (2)$
If You define...
$\displaystyle y_{1} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{0}\}\ d t\ (3)$
... and...
$\displaystyle y_{n} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{n-1}\}\ d t\ (4)$
... then the sequence of $y_{n}$ converges to the solution $y(x)$...
In Your case is $y_{0}=1$, $x_{0}=0$ and $y_{1} = (1 + x)^{2}$, so that is $f(x,1) = 2\ (1+x)$...
Kind regards
$\chi$ $\sigma$