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The integrating factor for this linear ordinary differential equation is $\mu (x)=e^{x^3/3}$. Multiplication gives an exact derivative:Consequently $y=1+C{e}^{-x^3/3}$. The initial condition $y(1)=2$ gives $C=e^{1/3}$, so the initial value problem has solutionThe integrating factor calculation determines every solution, and the initial condition fixes its remaining constant of integration uniquely.

Using the first partial derivatives from part (b),Also $uvg=yf$. Therefore the first partial differential equation becomesFor the second partial differential equation, the same chain rule givesAway from the origin, cancellation yields $f_x+2xf=0$. Treating $y$ as fixed, an integrating factor gives ${\partial }_x(e^{x^2}f)=0$. Thuswhere $F$ is an arbitrary sufficiently differentiable function. This is the local general solution, and the global form when $f$ is defined on the whole plane. If $f$ and its first partial derivatives are continuous at the origin, the reduced partial differential equation also holds there by continuity.

After division by $x^2$, the linear ordinary differential equation becomesThe integrating factor is $e^{1/x}$. ThusThe initial condition $y(1)=1$ gives $C=e+1$, so

With the change of variables $z=y^2$, multiplication of the differential equation by $2y$ gives $z^{\prime }+2xz=2x$. The integrating factor is $e^{x^2}$, soThe initial condition gives $C=3$. Since $y$ starts negative and $z\ge 1$, continuity selects the negative square root throughout:This never reaches the singular value $y=0$.

Divide by $x>1$. The product rule turns the linear ordinary differential equation intoEquivalently, $\ln x$ is an integrating factor after division by $x\ln x$. An antiderivative gives $\ln xy=\frac{1}{2}(\ln x{)}^2+C$. The initial condition at $e$ gives $C=-1/2$, henceIts derivative is $y^{\prime }=[1+(\ln x{)}^{-2}]/(2x)$, so substitution gives $x\ln x{y}^{\prime }+y=\ln x$, and $y(e)=0$. Every solution of a differential equation has the integrated form above, and the initial condition fixes its only constant. This agrees with the uniqueness theorem for ordinary differential equations on the stated interval.

Divide by $\cos x$ on an interval where it is nonzero. For the resulting first-order linear differential equation, use the integrating factor $M=\sec x+\tan x$, since $M^{\prime }=\sec xM$. The right-hand side after division is also $M^{\prime }$, soThe identity $(\sec x+\tan x)(\sec x-\tan x)=1$ shows $M$ never vanishes on the interval, givingAn integrating factor may have either fixed sign; if obtained by integrating $\sec x$, the real logarithm is $\ln |\sec x+\tan x|$.

Use integrating factor $e^{-x^3/3}$ for this first-order linear differential equation. Then $(y{e}^{-x^3/3}{)}^{\prime }=x^2{e}^{-x^3/3}=-(e^{-x^3/3}{)}^{\prime }$. Integrating givesThis includes the constant particular solution $y=-1$ when $C=0$.

On an interval with $\sin x\ne 0$, divide by $\sin x$ and use integrating factor $1/\sin x$. ThenPut $u=\sin x$ in the integral: $\int {\cos }^{3}xdx/{\sin }^{2}x=\int (u^{-2}-1)du=-u^{-1}-u+C$. Multiplying back givesAlthough the derivation divided by $\sin x$, this final expression is smooth and directly satisfies the original undivided equation at every real $x$, including multiples of $\pi$.

Factor the differential operator as $(D-a)(D-b)$, where $D=d/dx$. Its homogeneous solutions are $e^{ax},e^{bx}$, while a nonresonant particular solution is $e^{cx}/[(c-a)(c-b)]$. Imposing the two zero initial conditions gives the exponential forcing response with zero initial dataOne direct derivation avoids a large simultaneous system. Set $v=y^{\prime }-by$, so $v^{\prime }-av=e^{cx}$ and $v(0)=0$. The integrating factor gives $v=(e^{cx}-e^{ax})/(c-a)$; solving $y^{\prime }-by=v$ with $y(0)=0$ produces the displayed expression. This also verifies the zero derivative at the origin, since $v(0)=y(0)=0$.

For $x\ne 0$, divide by $x$ and use the integrating factor $\mu (x)=e^x|x{|}^{\alpha }$, separately on each side of zero. The equation becomes $(\mu y{)}^{\prime }=|x{|}^{\alpha }/x$. ThusFor $x>0$ the absolute values may be omitted. The original equation is singular as a first-order evolution equation at zero, so these formulas describe regular intervals with $x\ne 0$; extension through zero needs a separate regularity condition.

On the interval $x>-1$ containing the initial condition, division by $1+x$ gives a linear ordinary differential equation with integrating factor $(1+x{)}^{-3}$. ConsequentlyThe initial value gives $C=1$. ThusThis polynomial also satisfies the original equation at $x=-1$ and gives an extension to all real $x$. The initial value problem is uniquely determined on $x>-1$; the vanishing leading coefficient at $-1$ does not ensure unique continuation to its left.

Set $u=1/y$, so $u^{\prime }=-y^{\prime }/y^2$. The equation becomes $x(x^2-1)u^{\prime }+u=-x^3$. For $x>1$ its integrating factor is $\sqrt{x^2-1}/x$, givingThe initial value is $u(2)=-2$, forcing $C=0$. HenceSubstitution into the original equation verifies this answer throughout $x>0$, including $x=1$ where the divided linear equation was singular. Its maximal connected interval through the initial point is $(0,\infty )$.

Near the initial point, division by $\cos x$ gives $y^{\prime }+(1+\tan x)y=2\sec x+2\sin x$. An integrating factor is $e^x\sec x$, andThus $y=2\sin x+C{e}^{-x}\cos x$, and $y(0)=2$ givesAlthough division by $\cos x$ was local, direct substitution verifies this smooth solution in the original equation for every real $x$, including the zeros of $\cos x$.

Construct the scalar potential $F(x,y)=y{e}^{x^2}-x^2$. Its differential iswhich proves exactness directly. Multiplying the differential equation by $e^{x^2}$ shows that a solution satisfies $dF/dx=0$, so $y{e}^{x^2}-x^2=C$. The initial value at zero gives $C=-1$. ThereforeThe factor $e^{x^2}$ is also an integrating factor for the associated first-order linear equation.

First solve for the velocity components. The integrating factor $e^{\gamma t}$ givesIntegrating once more and imposing the initial position givesThese formulas describe projectile motion with linear drag. Differentiation checks both second-order differential equations and both initial velocities. At large positive time, the horizontal velocity tends to zero and the vertical velocity tends to the terminal value $-g/\gamma$.

The integrating factor is $e^{kx}$, giving $(e^{kx}y{)}^{\prime }=a{e}^{kx}\sin (mx)$. The required antiderivative isAfter dividing by $e^{kx}$, the general solution is the oscillatory term plus $C{e}^{-kx}$. The initial condition fixes $C=1+am/(k^2+m^2)$, soThe real, nonzero parameters ensure $k^2+m^2>0$; direct substitution verifies both the equation and $y(0)=1$.

For the first-order linear ordinary differential equation, an integrating factor isMultiplication converts the left side to a derivative:Integrating gives $(1+x^2{)}^{2}y=\arctan x+C$. The initial value fixes $C=1$, so

On a branch with $y\ne 0$, the given derivative is nonzero wherever the original equation is defined, so use inverse-function linearization of a first-order equation and regard $x$ as a function of $y$:The integrating factor $1/y$ givesThus the nonzero solutions are given implicitly byEach branch must also retain $x+y+2\ne 0$; at those excluded points the original right-hand side is undefined and the inverse graph cannot represent a regular $y(x)$. Differentiating the implicit equation gives $y^{\prime }=1/(\ln |y|+C+1)$, equal to the required quotient wherever these restrictions hold.

The additional solution $y=0$, on any interval avoiding $x=-2$, was lost when dividing by $y$ and must be included. It is not defined through the point $(x,y)=(-2,0)$ in the original equation.

For $y\ne 0$, the Bernoulli differential equation becomes linear under $u=1/y$. Multiplying the equation by $-y^{-2}$ givesOn an interval with $x\ne 0$, the integrating factor is $x^{-2}$, henceTherefore the complete set of solution branches isThe original equation requires $x\ne 0$; each nonzero branch additionally requires $C+\sin x\ne 0$. The zero solution is restored explicitly because the reciprocal substitution excludes it.

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.