Use the complex exponential representation of cosine. For real $\alpha$ and $\beta >0$, exponential decay makes the improper integral absolutely convergent, andFor the second integral, use differentiation under the integral sign with respect to $\beta$. The minus sign is essential: ${\partial }_{\beta }{e}^{-\beta x}=-x{e}^{-\beta x}$. HenceTo justify this step near any fixed ${\beta }_0>0$, restrict $\beta \ge {\beta }_0/2$. The absolute value of the differentiated integrand is at most $x{e}^{-{\beta }_{0}x/2}$, an integrable bound, so dominated convergence permits the differentiation. At $\alpha =0$ the answers reduce to $1/\beta$ and $1/{\beta }^2$, respectively.

Let $F(a)$ be the given integral. Its integrand extends continuously to $x=0$, with value $a$. The Leibniz integral rule includes both the moving upper endpoint and the derivative of the integrand. For $a\ne 0$ it givesConsequentlyThe zero case follows directly from $F(a)=a^3+O(a^9)$ near zero: expand $\sin (ax)/x=a+O(a^3{x}^2)$ and integrate up to $a^2$. Equivalently, the change of variables $t=ax$ gives $F(a)={\int }_0^{a^3}(\sin t)/tdt$, whose derivative is the same expression, interpreted continuously at zero. This substitution works for negative $a$ as an oriented integral too.

Take the dot product of the equation with $a$. Since $a\cdot (a\times r)=0$, we obtain $a\cdot r=(a\cdot c)/\lambda$. Next take its cross product with $a$ and apply the vector triple product identity:Substitute $a\times r=c-\lambda r$ and the known dot product, then rearrange:Here the vectors and $\lambda \ne 0$ are real, so the denominator is positive. The formula also covers $a=0$, giving $r=c/\lambda$. To check existence as well as uniqueness, write its numerator as $N$. The vector triple product identity gives $a\times N=\lambda a\times c-a(a\cdot c)+|a{|}^{2}c$, so $a\times N+\lambda N=({\lambda }^2+|a{|}^2)c$. Substitution verifies the original equation. This is an instance of the inverse of a scalar plus a cross-product operator.

Three vectors in three-dimensional Euclidean space are non-coplanar precisely when their scalar triple product is nonzero. Put $\Delta =a\cdot (b\times c)\ne 0$. In particular, $a\ne 0$. The identity for cross products with a common vector givesFor completeness, this follows from the vector triple product identity: writing $p=a\times b$ gives $p\times (a\times c)=a(p\cdot c)-c(p\cdot a)$, with $p\cdot c=\Delta$ and $p\cdot a=0$. Therefore the new scalar triple product isThus $a,a\times b,a\times c$ are non-coplanar, or equivalently linearly independent.

Write $\mathbf{r}=\mu \mathbf{n}+\mathbf{v}$. Since $\mathbf{n}$ is a unit vector, the equation becomes $\mathbf{v}\cdot \mathbf{n}=0$. Thus $\mathbf{v}$ ranges over the two-dimensional orthogonal complement of $\mathbf{n}$: the locus is an affine hyperplane through $\mu \mathbf{n}$, with unit normal $\mathbf{n}$. Conversely, every such vector satisfies the equation. Its perpendicular distance from the origin is $\boxed{|\mu |}$; $\mu$ is the signed distance in the chosen normal direction.

Integrate $f_x=y{e}^x$ with respect to $x$, obtaining $f=y{e}^x+g(y,z)$. Matching $f_y=e^x+2y{z}^3$ gives $g_y=2y{z}^3$, so $g=y^2{z}^3+h(z)$. Finally, $f_z=3y^2{z}^2+h^{\prime }(z)$ must equal the third component, so $h^{\prime }=0$. Thus a scalar potential isThe fundamental theorem for line integrals now gives ${\int }_{C}v\cdot dr=f(1,1,1)-f(0,0,0)=e+1$ for either path, confirming both direct computations. Since the curl of a gradient is zero for this smooth function, it also confirms the result in part (b). The potential is defined throughout ${\mathbb{R}}^3$, so no domain obstruction to path independence occurs.

Now the characteristic equation is $(m+1{)}^2=0$, with a repeated root, so the complementary function is $(C_1+C_{2}x)e^{-x}$. For the polynomial forcing, substitute $a{x}^2+bx+c$ to obtainThus $a=1$, $b=-4$, $c=6$. For the forcing $2e^{-x}$, both $e^{-x}$ and $x{e}^{-x}$ already occur in the complementary function; the resonant exponential particular solution therefore starts with $K{x}^2{e}^{-x}$. The identitygives $K=1$. The resulting general solution is

The characteristic equation is $m^2+3m+2=(m+1)(m+2)=0$, giving the complementary function $C_1{e}^{-x}+C_2{e}^{-2x}$. For the forcing term $x$, a linear particular solution $ax+b$ gives $2ax+3a+2b=x$, hence $a=1/2$, $b=-3/4$.

The forcing $-e^{-x}$ gives resonance in a differential equation, since $e^{-x}$ already appears in the complementary function. Try $Kx{e}^{-x}$. Direct differentiation givesso $K=-1$. Combining the two particular solutions givesThe factor $x$ in the resonant exponential particular solution is required because $-1$ is already a root of the characteristic equation.

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.

Recognize the left-hand side as a product rule derivative, and use $\frac{1}{2}\sin 2x=\sin x\cos x$:At $x=\pi /6$, both $y\sin x$ and $\frac{1}{2}{\sin }^{2}x$ equal $1/8$, so $C=0$. ThereforeDividing by $\sin x$ first obtains this solution on $(0,\pi )$, the interval containing the initial condition. The explicit expression is smooth on the entire real line and satisfies the original, undivided differential equation even where $\sin x=0$.

Separate the variables in this first-order ordinary differential equation:Integration gives $\frac{1}{2}\ln (y^2+2)=\frac{1}{2}\ln |x^2-3|+C$, so $y^2+2=K|x^2-3|$. On the interval containing the prescribed point $x=3$, we have $x^2-3>0$. The initial condition $y(3)=4$ therefore gives $18=6K$, hence $K=3$. Choosing the positive square root to match that initial condition yieldsThe maximal interval containing $3$ on which this solves the explicit ordinary differential equation is $\boxed{x>\sqrt{11/3}}$. At the left endpoint $y=0$, the original denominator vanishes and the derivative becomes unbounded. Directly, $y^{\prime }=3x/y$ and $x(2+y^2)/[y(x^2-3)]=3x/y$, verifying the equation throughout this interval.