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Collecting the shifted terms gives a second-order difference equation with coefficients $1-h/2$, $-(2+6h^2)$ and $1+h/2$. Substitute $y_n=r^n$ to obtain its characteristic equation of a linear recurrence:For real $h\ne 0,2$, the roots and general solution areIn the small-step regime, indeed for $0<h<2$, the polynomial is positive at $r=0$, negative at $r=1$ and positive for sufficiently large positive $r$. Thus $0<r_{-}<1<r_{+}$. The bounded solution of a second-order constant-coefficient recurrence discards the growing root, and $y_0=1$ fixes its remaining coefficient:The Taylor series establishes the displayed approximation to the root. Raising that approximation to the $n$th power is a leading approximation, rather than a uniform relative approximation for arbitrarily large $n$: its relative error is controlled when $n{h}^2$ is small. For completeness, the degenerate values give $y_n=A+Bn$ when $h=0$, and $y_n=C13^{-n}$ when $h=2$.

For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore giveHence $\mathbf{x}(t)=e^{tM}{\mathbf{x}}_0$ solves the initial value problem. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the matrix exponential into its even and odd powers provesThis is the matrix exponential when the square is minus the identity. If ${\mathbf{x}}_0=(x_0,y_0{)}^{\mathsf{T}}$, the general solution isThe constants in part (a) are $\alpha =x_0/5$ and $\beta =y_0+2x_0/5$, which makes the two forms identical.

Substituting the proposed form $\mathbf{v}(t)=e^{\lambda t}\mathbf{u}$ into the linear system of ordinary differential equations givesIf $\lambda$ is not an eigenvalue of $B$, the matrix $\lambda I-B$ is invertible, soprovides the required particular solution.

In the stated two-dimensional case,and therefore $\mathbf{u}=(3,2{)}^T$. The eigenvalues of $B$ are $\pm \sqrt{3}$, with corresponding eigenvectors $(\sqrt{3},1{)}^T$ and $(-\sqrt{3},1{)}^T$. Adding the general solution of the homogeneous system yields

For the given matrix $A$, the characteristic polynomial isFor $\lambda =4$, the eigenvector equation gives $v_2=v_1$; for $\lambda =-1$, it gives $3v_1+2v_2=0$. Thus the eigenvalues and their corresponding eigenspaces areFor any eigenvector $v$ with eigenvalue $\lambda$, the vector-valued function $e^{\lambda x}v$ satisfies $Y^{\prime }=AY$, since its derivative is $\lambda e^{\lambda x}v=e^{\lambda x}Av$. The two displayed eigenvectors are linearly independent, so the general solution of the linear system of ordinary differential equations is

The function $y_1=x$ solves the linear ordinary differential equation. Apply reduction of order by writing $y=xv(x)$. Substituting $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$ and cancelling the terms in $v$ givesFor $w=v^{\prime }$, this becomes a first-order linear ordinary differential equation:A partial fraction decomposition provides the required antiderivative:ThereforeThe logarithm has positive argument on the required interval. The Wronskian of $x$ and $1+x\log ((x-1)/x)$ is $1/(x-1)\ne 0$, proving their linear independence and hence completeness of this general solution.

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.

For $y=x$, the derivatives are $y^{\prime }=1$ and $y^{\prime \prime }=0$, and substitution gives $2(x-1)-2(x-1)=0$.

Use reduction of order: set $y=xv$, so $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$. Substitution and cancellation giveSince $x>0$, this reduces to $v^{\prime \prime }+2v^{\prime }=0$, hence $v^{\prime }=A{e}^{-2x}$ and $v=C_1+C_2{e}^{-2x}$. The general solution isThe two solutions are linearly independent, since their ratio $e^{-2x}$ is nonconstant; equivalently, their Wronskian is $-2x^2{e}^{-2x}\ne 0$.

The homogeneous linear differential equation $y^{\prime \prime }+y=0$ has general solution $C_1\cos x+C_2\sin x$. For a particular solution, use variation of parameters with $y_1=\cos x$, $y_2=\sin x$ and Wronskian one:The double-angle formulas and integration by parts giveSubstitution simplifies to $-x^2\cos x/4+x\sin x/4+\cos x/8$. Absorb the last term into the homogeneous linear differential equation solution. ThereforeIndeed, applying $d^2/d{x}^2+1$ to $x\sin x$ gives $2\cos x$, and applying it to $x^2\cos x$ gives $2\cos x-4x\sin x$; their difference divided by four is the required $x\sin x$.

For $y=x$, the first and second derivatives are one and zero, giving $-2x+2x=0$. To find an independent solution, use reduction of order and write $y=xv(x)$. Cancellation of the terms proportional to $v$ leavesWriting $w=v^{\prime }$ and using separation of variables givesSince $1/[x^2(1-x^2)]=x^{-2}+(1-x^2{)}^{-1}$, integration yields the general solutionThe second factor divided by $x$ has nonzero derivative $1/[x^2(1-x^2)]$, so the two solutions are linearly independent.

The product rule gives $y^{\prime }=x^n{z}^{\prime }+n{x}^{n-1}z$ and $y^{\prime \prime }=x^n{z}^{\prime \prime }+2n{x}^{n-1}{z}^{\prime }+n(n-1)x^{n-2}z$. On an interval avoiding $x=0$, substitution and division by $x^{n+2}$ giveFor a constant-coefficient differential equation, the coefficient of $z^{\prime }$ forces $2n+4=0$. At $n=-2$, the remaining coefficient $n^2+3n+2$ also vanishes. ThusThe characteristic equation of a constant-coefficient differential equation is $r^2+2r+2=0$, with complex conjugate roots of a polynomial $-1\pm i$. Its real solutions $e^{-x}\cos x$ and $e^{-x}\sin x$ are linearly independent. Therefore the general solution isThe constants can be chosen independently on the two sides of the singular point. The power substitution reducing a differential equation to constant coefficients here is invertible wherever $x\ne 0$.

For the homogeneous linear recurrence relation, try $u_n=r^n$. Its characteristic equation of a linear recurrence isThe distinct roots of a polynomial $1,-4$ give the complementary terms $A+B(-4{)}^n$. A constant cannot absorb the forcing because $1$ is already a root of a polynomial of the characteristic equation; try a particular solution $cn$. Substitution gives $c=-4c+1$, hence $c=1/5$. The general solution is $u_n=A+B(-4{)}^n+n/5$.

The initial conditions give $A+B=0$ and $A-4B=4/5$. Hence $A=4/25$, $B=-4/25$, and

The characteristic equation for the homogeneous solution is $(r+2{)}^2=0$, giving $(C_1+C_{2}x)e^{-2x}$. For a particular solution, substitute $A{e}^{2x}$: the left side becomes $16A{e}^{2x}$, so $A=1/16$. The general solution is

Set $\omega =\sqrt{k/m}$. Each coordinate satisfies the harmonic oscillator equation, so the general solution isThe four initial data give $A=X_0$, $B=C=0$, and $D=V_0/\omega$. ThusThis is the position component of the isotropic harmonic oscillator flow. Substitution also gives the constant energy $E=(k{X}_0^2+m{V}_0^2)/2$.

Using the sine addition formula, combine the surviving modes:It has amplitude $2$ and a phase delay $\pi /3$. The full general solution would add $C{e}^{-x/\sqrt{3}}$, but this exponential is periodic only for $C=0$, so the displayed periodic solution is unique.

Let $M=x+y$ and $N=x$. Their cross partial derivatives agree: $M_y=N_x=1$. More directly, construct the scalar functionThis exhibits the expression as an exact differential. Along a solution of the differential equation,Thus the general solution isFor $C=0$, the graph $y=-x/2$ also extends smoothly through $x=0$ and satisfies the original, undivided equation there. Nonzero $C$ cannot extend through zero as a finite-valued differentiable solution. The extra vertical component $x=0$ of the implicit zero-level equation is not a graph $y(x)$ on an interval and is not an additional solution of this ordinary differential equation.

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 $y(0)=1$.

The characteristic equation has the repeated root $r=-3$. Its homogeneous solution is therefore $(C_1+C_{2}x)e^{-3x}$. For a particular solution, try $y_p=Ax+B$. Substitution gives $9Ax+6A+9B=3x+1$, so $A=1/3$ and $B=-1/9$. Thus the general solution isThe extra factor $x$ in the second homogeneous term is required by the repeated root.

With $a=2$, integrate the $dx$ coefficient with respect to $x$:Its other partial derivative is ${\Phi }_y=x^2/2-2xy+g^{\prime }(y)$. Matching the $dy$ coefficient gives $g^{\prime }=0$. Therefore the differential equation is $d\Phi =0$, and its implicit general solution isThese are level curves of the potential of a conservative vector field. Where $N\ne 0$, they give the usual graph solution $dy/dx=-M/N$. At $N=0$, the implicit description retains possible vertical tangents instead of dividing by zero.

Multiply the separable differential equation by $e^y$: $e^y{y}^{\prime }=e^x$. The left side is $(e^y{)}^{\prime }$, hence $e^y=e^x+C$. ThereforeThe real general solution is taken on any connected interval where the logarithm's argument is positive.

The common homogeneous solution is $C_1\cos 2x+C_2\sin 2x$. For forcing $\sin x$, a trial $a\sin x$ gives $3a\sin x=\sin x$, so the first general solution isFor forcing $\sin 2x$, the frequency produces resonance in a differential equation. A trial $ax\cos 2x$ gives $y_p^{\prime \prime }+4y_p=-4a\sin 2x$, so $a=-1/4$ and the second solution isThe superposition principle permits adding the two particular solutions. The sum-forcing equation therefore hasOnly one pair of homogeneous constants is needed, because the two original homogeneous families coincide.