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Substitute $\mathbf{y}=\mathbf{v}{e}^{-\gamma t}$ into the unforced system. It gives $(K-\gamma I)\mathbf{v}=0$, so a nonzero single-exponential solution must use one of the eigendirections in part (a):These are all nonzero solutions having one common decay constant. Their arbitrary linear combination is the general unforced solution; all decay because the three eigenvalues are positive.

First divide each of the first two equations by two. ThenThe equal-component and opposite-component directions diagonalize the upper block, while the third coordinate is independent. Direct multiplication gives the eigenvalues and unit eigenvectorsThey form an orthonormal eigenbasis because $K$ is a real symmetric matrix.

Put $p=y^{\prime }$ and differentiate the general form of Clairaut's equation:Cancellation givesFor $f(p)=1/(p-1)$ the equation is defined only when $p\ne 1$. There are two types of smooth branch.

First, constant $p=C\ne 1$ gives the straight-line familySecond, $x+f^{\prime }(p)=0$ gives $x=1/(p-1{)}^2$, so $x>0$ and $p=1\pm x^{-1/2}$. Substituting in the original equation gives its two singular envelope branches,Their derivatives are indeed $1\pm 1/\sqrt{x}$, verifying the required value of $p$. There is no finite-slope continuation of either envelope through $x=0$, and no real envelope branch for $x<0$.

For completeness, if “solution” requires only a continuous first derivative, one can join a line from the family to the corresponding envelope at their tangency point $x=(C-1{)}^{-2}$. The function and its first derivative match there, so such piecewise branches also solve the original first-order equation. They are not twice continuously differentiable at a nontrivial join. The lines and envelope branches above exhaust the smooth branches; allowing these tangent joins supplies the additional $C^1$ solutions.

Write $P_k=P(X=k)$ for $k=0,1,2,\dots$, with ${\sum }_k{P}_k=1$. Provided the relevant moments exist, the expected value and variance areThe displayed support includes zero, so these are nonnegative integer values.

For $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is ${\mathbf{r}}^{\prime }(t)=(1,1,2t)$. Substitute into the field:Its scalar product with the tangent is $3t^2-4t-7t^6$. Hence the explicitly parametrized line integral isAs a check, the potential from part (d) has values $-128+C$ at $(2,2,4)$ and $-2+C$ at $(1,1,1)$, giving the same difference.

Evaluate the line integral along the three coordinate segments, treating coordinates not changing along a segment as constants. On the first segment, $y=z=0$, so $F_x=0$ and the contribution is zero. On the second segment,On the final segment,The fundamental theorem for line integrals therefore givesTaking its gradient verifies each component of the field. The constant is $C=\Phi (0,0,0)$; the field specifies only potential differences because the gradient of every constant is zero.