# IE2

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 $x\in [0,1]$.