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For this polynomial, $f^{\prime }(x)=2x$, so the specified Newton root-finding iteration simplifies toStarting at one gives $x_1=(1+2)/2=3/2$, followed by $x_2=(3/2+4/3)/2=17/12$. Consequently

Write the remaining quadratic polynomial as $x^2+bx+c$. Comparing coefficients in $(x-2)(x^2+bx+c)$ with the given cubic polynomial gives $b-2=-1$ and $-2c=-2$, hence $b=c=1$. The coefficient of $x$ is then $c-2b=-1$, as required. Thus

In spherical coordinates, integrating over the angular variables gives the volume element $4\pi r^{2}dr$ for a spherical shell. The mass density therefore gives the total massThus

Use the product-to-sum formula and the double-angle trigonometric identities:By Fourier orthogonality, these are already its only nonzero Fourier series coefficients:

This is a generalized eigenvalue problem. A nonzero vector lies in the kernel of $A-\lambda B$ precisely when its determinant vanishes:HenceThe corresponding eigenvectors can be taken as $(1,-1{)}^T$ and $(1,1{)}^T$, respectively, which also checks that both values give nontrivial solutions.

On an interval excluding $t=0$, an integrating factor for this first-order linear differential equation is $t^4$. Multiplying through givesAn antiderivative then gives $t^{4}y=3t^5/5+C$, soThe constant is arbitrary on each connected interval of the domain; the original differential equation is undefined at zero.