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The exponential function is strictly positive, so the zero level set is determined entirely by its prefactor:This is a rectangular hyperbola with branches in the first and third quadrants.

For real solutions with $x>0$, the square root requires $y\ge 0$. First retain the solution $y\equiv 0$, which would be lost by dividing by a square root of $y$.

For $y>0$, use the homogeneous first-order differential equation substitution $v=\sqrt{y/x}>0$, so $y=x{v}^2$, $\sqrt{xy}=xv$ and $y^{\prime }=v^2+2xv{v}^{\prime }$. The equation becomesThis is now a separable differential equation. Integration gives $-2/v+2\ln v=-\ln x+C$. Since $x{v}^2=y$, the positive solutions satisfyThe implicit family is sufficient here because only parts (a) and (b) request an explicit expression. It also shows why the nonnegative choice of $v$ matters: $\sqrt{xy}$ denotes the positive square root.

The original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the differential formThe cross partial derivatives of its coefficients are both four, so it is an exact differential form. Integrating its first coefficient in $x$ gives a potential $F=\frac{5}{2}{x}^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit general solution isDifferentiating verifies $y^{\prime }=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a derivative when $8y^3=4x$.

Rewrite the linear ordinary differential equation as $y^{\prime }+y=e^{3x}$. Multiplication by its integrating factor $e^x$ gives $(e^{x}y{)}^{\prime }=e^{4x}$. Integrating yieldsThe general solution consists of this particular solution and the homogeneous solution $C{e}^{-x}$.

For $x\ne 0$, divide by $x^2$ to obtain $y^{\prime }+y/x=1/x^2$, a first-order linear ordinary differential equation. An integrating factor is $x$ on either interval not crossing zero, since the product rule givesIntegration and division by $x$ give the explicit general solutionThe original equation would read $0=1$ at zero, so no solution can include that point.

By De Moivre's theorem, $\cos 5\theta$ is the real part of $(\cos \theta +i\sin \theta {)}^5$. Writing $c=\cos \theta$ and $s=\sin \theta$, the binomial theorem givesUse the Pythagorean trigonometric identity $s^2=1-c^2$ and expand:Therefore

Using the exponential definition of hyperbolic tangent, set $q=e^{2z}$:Here $q+1=1-i\ne 0$, so the resulting points are not poles. Taking all values of the complex logarithm gives $2z=i(-\pi /2+2\pi k)$. ThusEquivalently $\tanh (iu)=i\tan u$, so these are exactly the imaginary-axis solutions of $\tan u=-1$.

The factor $1+i$ has modulus $\sqrt{2}$ and principal argument $\pi /4$. Multiplication by $e^{i\pi /6}$ adds $\pi /6$ to its complex argument, while the real factor $1/2$ halves its modulus. HenceTaking the principal logarithm for $w$ givesThis complex number lies in the second quadrant. Its modulus follows from the sum of squared components, and its principal argument follows by the second-quadrant correction:All arguments of this fixed $w$ add integer multiples of $2\pi$ to the displayed one.

If $\ln z$ is interpreted as the multivalued complex logarithm, there is instead a family $w_k=-a+i{b}_k$, where $a=\frac{1}{2}\ln 2>0$ and $b_k=5\pi /12+2\pi k$. Then $|w_k|=\sqrt{a^2+b_k^2}$, and the principal argument is $\pi -\arctan (b_k/a)$ for $b_k>0$, or $-\pi -\arctan (b_k/a)$ for $b_k<0$. This records the logarithm-branch choice that the paper leaves implicit.

The two derivatives are $3x^2+3$ and $6x$, both equal to six at $x=1$. Their common tangent line through $(1,4)$ is $y-4=6(x-1)$, or $y=6x-2$. Setting $x=0$ gives its intersection with the vertical axis:

Equating the two polynomials gives $x^3-3x^2+3x-1=(x-1{)}^3=0$. Thus the only intersection has $x=1$ and $y=4$:

Substitution in the logarithmic expression gives $x\simeq 23.8/0.301=79.069\dots$. To two significant figures,

Take the base-ten logarithm of both positive sides. The exponentiation rule for logarithms gives $x{\log }_{10}2=y$, hence

The same chain rule gives $u_{xx}=u^{\prime \prime }(\xi )$ and $u_y=u^{\prime }(\xi )/\sqrt{y}$. Substituting these and part (a) into the partial differential equation givesThus every twice-differentiable profile $u(\xi )$ gives a solution on $y>0$.

For real coordinates, work on $y>0$. The chain rule gives ${\xi }_y=y^{-1/2}$ and ${\xi }_{yy}=-\frac{1}{2}{y}^{-3/2}$. ThereforeThe derivatives on the right are with respect to $\xi$. Behaviour at $y=0$ requires separate regularity of the composed function.

Here $y=0$, the unit normal vector is ${\mathbf{e}}_y$, and $\mathbf{F}\cdot {\mathbf{e}}_y=x^2$. The square has $-1/2\le x,z\le 1/2$, so its outward flux is