IE2

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Approximate $y(x) = x^2 + \int_0^1 \sin(xz) y(z) dz\,$

One method: Expand both $\,y(x)$ and $\,\sin(xz)$ in Taylor series; $\,y$ in powers of $\,x$ , and $\,\sin(xz)$ in powers of $\,xz$ . We can save ourselves some effort if we note from the integral equation that $\,y(0)=0$ ; and from repeatedly differentiating the integral equation, that $\,y''(0)=2\,$ (so the coefficient of the $x 2$ term in the $y$ expansion is 1), and that $\,y^{(2n)}(0) = 0$ for $n=2,3,\dots\,$ (because we get the sine in the integrand for those terms). Thus, the Taylor series expansion for $\,y(x)$ (about $\,x=0$ ) has the form $\,y(x) = c_1\,x + x^2 + c_2\,x^3 + c_3\,x^5 + \dots$ .

The expansion for $\,\sin(xz)$ is $\sin(xz) = xz - (xz)^3/3! + (xz)^5/5! +\dots\,$ .

Inserting these in the integral equation, we get $\,c_1 x + x^2 + c_2 x^3 + c_3 x^5 + \dots = x^2 + \int_0^1 (xz - \frac{x^3 z^3}{6} + \frac{x^5 z^5}{120} + \dots) (c_1 z + z^2 + c_2 z^3 + c_3 z^5 + \dots)\,dz$

or after some expanding and rearrangement (and subtracting the $\,x^2$ term from both sides)

$\,c_1 x + c_2 x^3 + c_3 x^5 + \dots = x\int_0^1(c_1 z^2 + z^3 + c_2 z^4 + c_3 z^6 + \dots)\,dz\, - \frac{x^3}{6}\int_0^1 (c_1 z^4 + z^5 + c_2 z^6 + c_3 z^8 + \dots)\,dz$
$+ \frac{x^5}{120}\int_0^1 (c_1 z^6 + z^7 + c_2 z^8 + c_3 z^{10} + \dots)\,dz -+\dots$ .

All of the integrals on the right side are readily evaluated. For example, retaining only terms through $\,x^5$ , we get

$\,c_1 x + c_2 x^3 + c_3 x^5 = x(\frac{c_1}{3} + \frac{1}{4} + \frac{c_2}{5} + \frac{c_3}{7})\, - x^3 (\frac{c_1}{30} + \frac{1}{36} + \frac{c_2}{42} + \frac{c_3}{54}) + x^5 (\frac{c_1}{840} + \frac{1}{960} + \frac{c_2}{1080} + \frac{c_3}{1320})$

Equating coefficients for each power of $\,x$ on the left and right sides gives a system of equations for the coefficients $\,c_1, c_2, c_3.$ Upon solving that system, we get $y(x)\approx 0.36361\,x + x^2 - 0.038996\,x^3 + 0.0014395\,x^5$ . Repeating the calculation, this time including terms through $\,x^7$ , gives $c_4 = -2.7175\times 10^{-5}\,$ and leaves the lower-order coefficients essentially unchanged, evidence that the above approximation for $\,y(x)$ is quite good for $x\in[0,1]$ .

Main Page : Integral Equations

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