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Apply integration by parts with $u=\log x$ and $dv=xdx$. Since $du=dx/x$ and $v=x^2/2$,The original PDF specifies this integral; the local OCR replaces the entire question by unrelated text.

Use the Maclaurin series $\sin 3x=3x-\frac{9}{2}{x}^3+O(x^5)$ and $\cos x=1-\frac{1}{2}{x}^2+O(x^4)$. Multiplication givesThe same terms follow from the two sine harmonics in the Fourier series found above.

Let $V$ be the enclosed region with piecewise smooth closed boundary $S$, oriented by its outward unit normal $\mathbf{n}$. For a continuously differentiable vector field on a neighbourhood of $V$, the divergence theorem statesThe outward flux integral equals the volume integral of the divergence.

The product-to-sum formula gives the finite Fourier seriesThus the only nonzero Fourier coefficients are $b_2=b_4=1/2$.

The chain rule gives $u_{xx}=f^{\prime \prime }(x+\lambda y)$ and $u_{yy}={\lambda }^2{f}^{\prime \prime }(x+\lambda y)$. The partial differential equation becomesFor this to hold for every twice differentiable $f$, the coefficient must vanish:For an affine $f$ alone any $\lambda$ would work, but the requirement is for arbitrary $f$.

The coefficient matrix has determinant $-1$, so it is an invertible matrix. Its homogeneous linear system therefore has only the zero solution:One can also subtract the first equation from the second to get $z=0$, then subtract the first from the third to get $y+2z=0$, giving $y=0$ and finally $x=0$. Thus there are no nonzero solutions.

The equation is not separable, because $x^2-xy=x(x-y)$ cannot be written as a product of a function of $x$ and a function of $y$ on an open region. For example, the ratio $(x^2-x{y}_1)/(x^2-x{y}_2)=(x-y_1)/(x-y_2)$ depends on $x$ for distinct fixed $y_1,y_2$, whereas a separated product would give a constant ratio.

It is not homogeneous in the sense of a homogeneous first-order differential equation: the right-hand side is not a function of $y/x$ alone. It is also not a homogeneous linear differential equation, since moving the $xy$ term to the left leaves a nonzero forcing.

It is linear, since it has the standard linear ordinary differential equation formThus the classifications are (a) no, (b) no, (c) yes.