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For $p$ in a half-plane of absolute convergence, interchange the integrals using Fubini's theorem:For positive real $p$, integration by parts gives $I_n={\int }_0^{\infty }{e}^{-pt}{t}^{n}dt=n{I}_{n-1}/p$, starting from $I_0=1/p$. Hence $I_n=n!/p^{n+1}$, orThe formula extends throughout $\mathrm{Re}p>0$ by the same integration argument. Write $X=\mathcal{L}x$. The integral term is the convolution of $t$ with $x(t)$, whose Laplace transform is $X/p^2$. The derivative transforms to $pX-x(0)=pX-1$. Thereforeand elementary factorization givesInverting the two polynomial transforms yieldsDirect substitution verifies $x^{\prime }=t$ and ${\int }_0^t(t-s)(1+s^2/2)ds=t^2/2+t^4/24$, as required. Uniqueness also follows without transform assumptions: differentiating the equation twice gives $x^{(3)}+x=1+t^2/2$ with initial data $x(0)=1$, $x^{\prime }(0)=0$, $x^{\prime \prime }(0)=1$, a linear initial-value problem with a unique solution.

Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the autonomous differential equation makes its trajectory a simple closed curve $C$, bounding a region $D$ by the Jordan curve theorem. The Green theorem givesAlong the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the Bendixson-Dulac criterion in this case.

For the second-order equation set $y=x^{\prime }$, giving $x^{\prime }=y$, $y^{\prime }=-y-\epsilon y^3-\sin x$. Choose the Dulac function $\rho =-1$. ThenThus there is no nonconstant periodic solution for $\epsilon >0$. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H^{\prime }=-y^2-\epsilon y^4$, whose integral over a period could vanish only if $y\equiv 0$.

The product $(\tanh s)y(s)$ is continuous, so the fundamental theorem of calculus makes the integral continuously differentiable. Differentiating the integral equation and then evaluating at zero gives $y^{\prime }=(\tanh x)y$, $y(0)=1$. Since $(\log \cosh x{)}^{\prime }=\tanh x$, an integrating factor is $1/\cosh x$. Thus $(y/\cosh x{)}^{\prime }=0$, andPut $L(x)=\log \cosh x$, so $L(0)=0$ and $L^{\prime }=\tanh x$. The claimed Picard iterate formula holds for $n=0$. Assuming it at $n$, termwise integration of the finite sum givesThis proves by inductionThe same calculation works for negative $x$ because it uses an antiderivative rather than a monotonicity substitution. On every bounded interval $|x|\le R$, $0\le L(x)\le \log \cosh R$, so comparison with the exponential-series tail also proves uniform convergence there.