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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].

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