Write the sine argument as $\pi /2+\delta$, with $\delta =\frac{\pi }{2}(e^x-1)=\frac{\pi }{2}x+\frac{\pi }{4}{x}^2+O(x^3)$. The sine addition formula gives $\sin (\pi /2+\delta )=\cos \delta$. Its Maclaurin series is $1-{\delta }^2/2+O({\delta }^4)$, soThere is no linear term because the outer sine has zero derivative at $\pi /2$.

Equal energies mean $dE=0$ in the linear approximation. Evaluate the coefficients at particle $A$, where $v_A=0.90c$, and use $\Delta v=0.01c$. The fractional rest mass change isConsequentlyAs a check beyond the requested first-order approximation, exact equality of relativistic energy gives $m_B/m_A=\sqrt{(1-0.91^2)/(1-0.90^2)}=\sqrt{1719/1900}\approx 0.95118$, consistent with the sign and approximate size of the difference.

Take the total differential of the logarithm of the relativistic energy, with the speed of light fixed:Using the chain rule,Every term is a dimensionless fractional change. The differential relation is exact; applying it to finite small changes is a first-order Taylor expansion.

The real equation has domain $x>0$. Its integrating factor isThus the linear ordinary differential equation becomes $(x^{x}y{)}^{\prime }=x^x{x}^{-x}=1$. After integration, $x^{x}y=x+C$. The initial condition gives $2=1+C$, hence

For $x\ne 0$, multiply by $x$ to obtain the first-order linear ordinary differential equation $y^{\prime }+xy=5x{e}^{-x^2}$. Its integrating factor is $\mu =e^{x^2/2}$. The product rule then givesIntegration yields $y{e}^{x^2/2}=-5e^{-x^2/2}+C$, soThe constant of integration is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains $1/x$ and is not defined there.

The first boundary condition requires $u(x,0)=a{x}^2+d=0$ for all $x$, so $a=d=0$. Since $u_y=bx+2cy$, the second boundary condition gives $b=0$. The partial differential equation then reads $2a-2c=1$, fixing $c=-1/2$. Thus

The chain rule for the new coordinates gives ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ and ${\partial }_y={\partial }_{\xi }-{\partial }_{\eta }$. Since the coordinates depend linearly on $x,y$, their coefficients are constant. Expanding the squares of these differential operators givesHence the partial differential equation becomes $u_{\xi \eta }=1/4$. Integrating once in each variable introduces independent one-variable functions, giving the general solutionwhere $F,G$ are arbitrary twice-differentiable functions on the relevant characteristic intervals. Their sum includes an arbitrary additive constant.

Eliminate $z$ using the constraint: $z=x-x^2-y^2$. The restricted objective is the quadratic polynomial $F(x,y)=axy+x-x^2-y^2$, with constant Hessian matrixIts eigenvalues are $-2+a$ and $-2-a$. HenceFor $|a|<2$, both eigenvalues are negative and the quadratic polynomial is strictly concave, so the maximum is global on the entire constraint surface. For $|a|>2$, the opposite signs give increasing and decreasing directions. At $a=\pm 2$ there is no stationary point, as already found, rather than a degenerate extremum to classify. No parameter value produces a constrained local minimum.

Let $h=x^2+y^2+z-x$. Using Lagrange multipliers with $\nabla f=\lambda \nabla h$ givesThe third equation fixes $\lambda =1$. The first two then reduce to $2x-ay=1$ and $2y=ax$. If $a^2\ne 4$, their solution and the constraint giveAt $a=2$, the second equation requires $y=x$ and makes the first inconsistent; at $a=-2$, it requires $y=-x$ and gives the same inconsistency. Thus there is no constrained stationary point when $a=\pm 2$. The constraint gradient never vanishes because its vertical component is one, so the multiplier condition has not excluded any regular constrained stationary points.

At $\mu =-1$, the first equation gives $x=y$, and the second then gives $z=0$. All nonzero solutions areAt $\mu =6$, the first equation gives $x=-6y$. The second then gives $3z=7y$, and the third is twice the second. Choosing an integer direction vector, all nonzero solutions areEach full null space, including the zero solution, is a line through the origin. Geometrically, the three homogeneous constraint planes have that common intersection; in each case only two of their equations are independent.

Subtract twice the second row from the third row when computing the determinant. The last row becomes $(0,0,\mu -6)$, soIts zeros give exactly the parameter values with a nontrivial null space:

For $\mathbf{b}=0$, a nonzero coefficient vector $(x,y,z)$ with $x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3=0$ is precisely a linear dependence of the three columns. Three vectors in three dimensions are linearly dependent exactly when the determinant of their column matrix, or their scalar triple product, is zero. ThusFor this square homogeneous linear system, the condition is also sufficient.

Take the outward orientation of the complete closed surface, including its flat base. The constant part of the vector field has zero divergence, and the radial part has divergence $3\mathbf{a}\cdot \mathbf{b}$. The divergence theorem gives its flux as that constant times the solid hemisphere's volume, $2\pi R^3/3$:The constant field may have nonzero flux through the curved part alone, but its curved and base contributions cancel on the closed surface.

Use the orientation induced by the boundary order $O\to A\to B\to O$. The triangle's vector area element integrates to $\frac{1}{2}\mathbf{a}\times \mathbf{b}$. Every point $\mathbf{r}$ in its plane is a linear combination of $\mathbf{a},\mathbf{b}$, so $\mathbf{r}\cdot (\mathbf{a}\times \mathbf{b})=0$. The radial term of the field therefore contributes no flux. The constant term givesReversing the orientation reverses this sign. If $\mathbf{a}$ and $\mathbf{b}$ are collinear, the triangle has zero area and the displayed value is zero by the degenerate-area convention.

Again write the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. The second term has zero curl. Applying the product rule to the first gives $\nabla (\mathbf{a}\cdot \mathbf{r})\times \mathbf{b}=\mathbf{a}\times \mathbf{b}$. Thus

The vector triple product rewrites the field as $\mathbf{b}(\mathbf{a}\cdot \mathbf{r})-\mathbf{r}(\mathbf{a}\cdot \mathbf{b})$. Its first term has divergence $\mathbf{b}\cdot \nabla (\mathbf{a}\cdot \mathbf{r})=\mathbf{a}\cdot \mathbf{b}$, while the second contributes $3\mathbf{a}\cdot \mathbf{b}$. Therefore

The curl of the constant vector field $\mathbf{a}\times \mathbf{b}$ vanishes. The curl of the position field $\mathbf{r}=(x,y,z)$ also vanishes, since all its cross-coordinate partial derivatives are zero. Thus

The fixed vectors have constant components, so $\mathbf{a}\cdot \mathbf{b}$ is constant, $\nabla \cdot \mathbf{a}=0$ and $\nabla \cdot \mathbf{r}=3$. Hence the divergence is

The characteristic polynomial factors as $(r+1)(r+2)$, so the homogeneous solution is $A{e}^{-x}+B{e}^{-2x}$. The forcing $e^{-x}$ already occurs in that solution, so a constant multiple of $e^{-x}$ cannot be a particular solution. Use $y_p=Cx{e}^{-x}$ instead. Its derivatives are $C(1-x)e^{-x}$ and $C(x-2)e^{-x}$, so substitution gives $C{e}^{-x}$ and fixes $C=1$. ThusThe extra factor of $x$ accounts for the repeated forcing exponent in the method of undetermined coefficients.

For the homogeneous linear differential equation, the characteristic equation is $r^2-2r+2=0$, with polynomial roots $1\pm i$. Its homogeneous solution is therefore $e^x(A\cos x+B\sin x)$.

Use the method of undetermined coefficients for a particular solution $y_p=a{x}^2+bx+c$. Substitution givesEquating coefficients yields $a=1$, $b=2$ and $c=1$. Hence the general solution is