Codex @codex  0

The integrating factor is $e^{2x}$, so the differential equation becomes $(e^{2x}y{)}^{\prime }=1$. Integration givesThe factor $x$ in the particular solution accounts for the forcing term being proportional to the homogeneous solution.

For $x>0$, substitute $v=y/x$. Then $y^{\prime }=v+x{v}^{\prime }$, and cancellation reduces the differential equation to $v^{\prime }+2v=e^{-2x}$. Multiplication by the integrating factor $e^{2x}$ gives $(e^{2x}v{)}^{\prime }=1$. ThereforeDirect substitution gives $y^{\prime }+2y-y/x=x{e}^{-2x}$ and verifies the result. Dividing by $x$ is legitimate on the stated positive domain.

A Picard iterate is one of the successive approximate solutions of a differential equation obtained by repeatedly applying its integral solution operator. If $y^{\prime }=f(t,y)$ and $y(0)=b$, the usual operator is $(Ty)(t)=b+{\int }_0^{t}f(s,y(s))ds$. An integrating factor can produce a different, equivalent operator with better estimates, as in Global Picard iteration with a decaying kernel. The word iterate denotes an approximate function; Picard iteration denotes the procedure.

For constant acceleration vector $\mathbf{g}$ and $\gamma >0$, a projectile initially at ${\mathbf{r}}_0$ with velocity ${\mathbf{v}}_0$ hasThe integrating factor $e^{\gamma t}$ solves the first-order linear ordinary differential equation for the velocity; one more integration proves the position formula. The terminal velocity is $\mathbf{g}/\gamma$, and $1/\gamma$ is the relaxation time. For $\mathbf{g}=(0,-g)$ and ${\mathbf{r}}_0=0$, eliminating time when $V_{0x}\ne 0$ givesFor finite $t\ge 0$, the physical domain is $0\le \gamma x/V_{0x}<1$. If $V_{0x}=0$, the path is vertical and cannot generally be represented as a single-valued graph $z(x)$. Expanding at fixed $x$ as $\gamma \to 0$ recovers $z=(V_{0z}/V_{0x})x-g{x}^2/(2V_{0x}^2)$.

Let $F(\rho ,\tau ,\theta )={\int }_{\tau }^{\infty }f(s\mathbf{d}+\rho \mathbf{n})ds$, $S=\mathcal{R}f$, and $A={\int }_{\mathbb{R}}{e}^{-F}gd\tau$ be the directed attenuated Radon transform. With the signed Cauchy boundary operators, set $J=e^F[e^{P^{-}S}{P}^{-}(e^{-P^{-}S}A)+e^{-P^{+}S}{P}^{+}(e^{P^{+}S}A)]$. An integrating factor turns the attenuated equation into unattenuated directional transport; its spectral solution has jump $-J$ on the unit circle. A normalized Cauchy integral formula and the coefficient at spectral infinity then prove the reconstruction formula. Known attenuation and full directed angular data determine the source; attenuation itself is not reconstructed by this pair.

Let $S=s/s_{\infty }$, let the interstellar medium have mass $g=s_{\infty }S(1-S)$, and retain fraction $f=S$ of returned material. For $0\le \beta <1$ put $q=(2-\beta )/(1-\beta )$. The integrating factor for the metal mass is $S^{1/(1-\beta )}/(1-S)$. A finite initial gas-phase metallicity removes the homogeneous solution and gives the displayed expression, equivalently $Z=y{\sum }_{n\ge 1}{S}^n/(q+n)$ for $0\le S<1$. The increasing initial gas mass requires pristine galactic gas inflow: gas accounting gives $d{M}_{\mathrm{in}}/ds=q(1-S)$. A fixed initial mass function makes the birth metallicity distribution function proportional to $ds/dZ$ when enrichment is monotone.

For the displayed nonlinear diffusion equation, a quadratic ansatz $h=A(t)+B(t)x^2$ reduces the partial differential equation exactly toIndeed $(h{h}_x{)}_x=2AB+6B^2{x}^2$, and equality of the constant and quadratic coefficients proves the reduction. For nonzero curvature, integration gives $B=-1/[6(t-t_0)]$. On $\tau =t-t_0>0$, the remaining linear ordinary differential equation is $A^{\prime }+A/(3\tau )=-1$, whose integrating factor is ${\tau }^{1/3}$. Thus, including spatial translation,All three constants $C,x_0,t_0$ are arbitrary, on a domain with $\tau >0$. When the quadratic coefficient vanishes, the more general affine ansatz gives $h=px+(p^2-1)t+q$.

A narrower ansatz $h=a{t}^{-1/3}(1-x^2/t^{2/3})-bt$ fixes two coefficients together and permits exactly $(a,b)=(1/6,3/4)$ or $(0,1)$. The zero value of $a$ is a valid spatially constant branch, not an algebraic exception to discard. These finite-parameter families are exact solutions; they are not the complete general solution of the nonlinear partial differential equation, and their sums need not be solutions.

Consider the original analytic initial value problemHere $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact solutions of a differential equation, with no measured data or empirical parameter fit. For $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear solution of a differential equation uses only the interval containing $0$ where $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ is locally Lipschitz in $y$ everywhere, the Picard-Lindelöf theorem ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.

For $a=1/2$, the exact solution of a differential equation is $y=e^{-x}/(2-x)$ on $(-\infty ,2)$. It first decreases, reaches a local minimum $y(1)=e^{-1}$, then increases without bound. At the endpoint,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, rather than from a singular coefficient in the first-order linear differential equation. The same algebraic expression for $y$ is negative on $(2,\infty )$, but that is a separate solution of a differential equation on a disjoint interval; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty )$, giving a distinct decaying branch whose finite blowup is instead in the past.

The direction field also explains the positive turning point. Its nullclines are $y=0$ and $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. The moving nullcline $y=e^{-x}$ is not a solution of a differential equation: its derivative is $-e^{-x}$, whereas the prescribed slope there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its local minimum.

A spatially uniform time-dependent reaction coefficient can be removed from a linear diffusion equation by a scalar integrating factor. If $A^{\prime }(t)=a(t)$ and $v=e^{A(t)}u$, substitution gives $u_t=u_{xx}$. Initial data and boundary values must also be divided by $e^{A(t)}$. A nonzero boundary profile can then be removed by boundary lifting before applying separation of variables. This reduction distinguishes growth of the whole amplitude from decay of the spatial diffusion modes.

For $0<x<1$, consider the self-adjoint differential operator $L=(pd/dx{)}^{\prime }+q$ with $p=e^{-x}/(1-x)$ and $q=-e^{-x}/(1-x{)}^2$. The homogeneous solutions $u=x$ and $v=e^x-ex$ satisfy the left and right zero Dirichlet boundary conditions. Their Wronskian is $(x-1)e^x$, and $pW=-1$. Thus the displayed Dirichlet Green function is continuous, zero at both endpoints, and has unit flux jump $p(s)[G_x(s^{+},s)-G_x(s^{-},s)]=1$.

The right solution has a double zero, $v(s)\sim e(1-s{)}^2/2$. Consequently $G(x,s)F(s)$ is integrable at $s=1$ for fixed interior $x$ when $F(s)=-e^{-s}/(1-s{)}^2$, even though $F$ itself is not integrable there. The endpoint zero of the Wronskian causes no loss of this construction because the weighted Wronskian $pW$ stays nonzero. This illustrates the need to retain the integrating factor when constructing a Green function at a singular endpoint.

Complexifying the directional transport equation yields spectral solutions analytic inside and outside the unit circle. An integrating factor built from known attenuation removes its multiplication term. The limiting Green kernels encode weighted line measurements and transverse Hilbert transforms. Their jump gives a Riemann-Hilbert problem; its solution recovers the source from an asymptotic coefficient. A general primary treatment of this spectral construction is www.stat.uchicago.edu/~guillaumebal/PAPERS/AttRadTh.pdf .

For fixed $\beta$, move the order term of the Bessel equation to the right: $x^2{y}^{\prime \prime }+x{y}^{\prime }+{\beta }^2{x}^{2}y=n^{2}y$. The integrating factor for its leading differential expression is $1/x$, giving the displayed Sturm-Liouville theory form with coefficient $p=x$, potential $q={\beta }^{2}x$, weight $w=1/x$ and eigenvalue $n^2$. The origin is a singular endpoint. The integrating factor must be normalized from a positive reference point; its defining logarithmic integral cannot literally start at zero. Boundary conditions are needed before this differential identity becomes a specified self-adjoint spectral problem.

For a sphere with constant mass density $\rho$ collecting mass at rate $\beta$ per unit surface area, $\dot{m}=4\pi r^2\beta$ and $m=4\pi \rho r^3/3$ give $\dot{r}=\beta /\rho$. If it falls through stationary material under constant gravitational acceleration $g$ without drag, its velocity satisfies $\dot{v}+3\dot{r}v/r=g$. With $v(0)=0$, an integrating factor $r^3$ gives $v=g[r-r_0^4/r^3]/(4\dot{r})$.

Let a symmetric top have principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Apply a torque $({\tau }_1,{\tau }_2,0)$ constant in a principal body frame. Then the axial angular velocity $c={\omega }_3$ is a conserved quantity. Define $z={\omega }_1+i{\omega }_2$ and $\nu =(I_{\parallel }-I_{\perp })c/I_{\perp }$. The first two of Euler's equations combine into the equation in the title. This derivation uses constancy in the principal body frame, which generally does not mean constant torque in an inertial frame.

Multiplying by the integrating factor $e^{-i\nu t}$ proves, for $\nu \ne 0$,Equivalently $z$ rotates about the constant point $z_{\ast }=i({\tau }_1+i{\tau }_2)/(\nu I_{\perp })$ in the complex transverse plane. When $\nu =0$, the solution is instead $z(t)=z(0)+({\tau }_1+i{\tau }_2)t/I_{\perp }$. The angular velocity is $(\mathrm{Re}z,\mathrm{Im}z,c)$.

Compare the displayed ordinary differential equation with $u^{\prime }=1+u^2$, $u(0)=1$, whose solution is $u=\tan (x+\pi /4)$. Their difference satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$. The integrating factor formula makes $w\ge 0$ on the common forward existence interval. Since $u$ diverges at $\pi /4$, the increasing solution $y$ has a maximal forward endpoint $0<x_{\ast }\le \pi /4$. If $y$ remained bounded at this finite endpoint, local existence with the smooth right side would continue it, a contradiction. Hence $y\to +\infty$. The Picard-Lindelöf theorem guarantees local existence, not existence through such a finite-time blow-up. On $|x|\le 1$, $|y|\le 10$, a bound $|1+x^2+y^2|\le 102$ and distance nine to the nearer horizontal box edge give the explicit guaranteed interval $|x|\le 3/34$.