For $h=x^2/a^2+y^2/b^2+z^2-1$, the gradient $\nabla h=2(x/a^2,y/b^2,z)$ is a normal vector to the ellipsoid. The outward unit normal vector is $(x/a^2,y/b^2,z)/\sqrt{x^2/a^4+y^2/b^4+z^2}$. The conditions $a,b>\sqrt{2}$ ensure that $z=\sqrt{1-x^2/a^2-y^2/b^2}>0$ everywhere on the rectangular projection.

The dot product $F\cdot \nabla h=2xy(1/b^2-1/a^2)$ vanishes pointwise when $a=b$. In that case $F$ has no normal component, so the surface flux is zero. For unequal axes, the given vector surface element givesPut $u=x^2/a^2$, $v=y^2/b^2$. Then $xydxdy=a^2{b}^{2}dudv/4$, andTwo elementary integrations now giveThis is the rotational flux through an ellipsoidal graph patch with unit vertical semiaxis and unit projection lengths. It also tends continuously to zero when $a=b$, agreeing with the pointwise argument.

Use Gaussian elimination. The last equation gives $y=c-z$, and the second then gives $x=b-4c-2z$. Substitute these into the first equation to obtain $3b-10c-z=a$. Back-substitution yieldsThe linear system has matrix formThe coefficients of the arbitrary right-hand side in the solved expressions are the rows of the matrix inverse. ThereforeMultiplying the original matrix by this candidate gives the identity matrix; its determinant is $-1$, so the solution is unique for every $(a,b,c)$.

For an odd integrable function, both $f(x)$ and $f(x)\cos nx$ are odd. Their integrals over $[-\pi ,\pi ]$ therefore vanish, so the constant and cosine Fourier coefficients are zero. For $f(x)=\sin ax$, the remaining Fourier sine coefficients areThe product-to-sum identity in the first step reduces the integral to two cosines. Here $a$ is real and noninteger, so no denominator vanishes. Consequently the Fourier series of a noninteger-frequency sine isAt $x=\pm \pi$ every term is zero. The Fourier series converges there to the average of the two limits of the periodic extension, namely zero, rather than to the nonzero endpoint values $\pm \sin \pi a$.

Use the method of undetermined coefficients. Substitution of $y_p=(a+bx)\sin x+(c+dx)\cos x$ givesEquating the coefficients of $x\sin x$, $x\cos x$, $\sin x$, and $\cos x$ to the forcing gives $d=1$, $b=0$, $c=1$, $a=-1$. ThusThe characteristic equation of the homogeneous equation is $(r-1{)}^2=0$, with a repeated root. Its complementary function is $(C_1+C_{2}x)e^x$. Since $y_p(0)=1$ and $y_p^{\prime }(0)=0$, the initial conditions give $C_1=-1$ and $C_2=1$. ThereforeDirect differentiation verifies both zero initial data and the original forcing.