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Both functions satisfy the constant-coefficient homogeneous linear differential equationIts characteristic polynomial is $\lambda ({\lambda }^2+1)$, so the general real solution is $y=A+B\cos x+C\sin x$. The three initial conditions give $A+B=0$, $C=1$, and $-B=0$, henceHere the equation has continuous coefficients and nonzero highest-derivative coefficient everywhere. The uniqueness theorem for ordinary differential equations, applied to its three-component first-order system, gives a unique solution on the entire half-line.

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 the homogeneous linear differential equation, the characteristic equation is $r^2-2r+2=0$, with polynomial roots $1\pm i$. Its homogeneous solution is therefore $e^x(A\cos x+B\sin x)$.

Use the method of undetermined coefficients for a particular solution $y_p=a{x}^2+bx+c$. Substitution givesEquating coefficients yields $a=1$, $b=2$ and $c=1$. Hence the general solution is

The characteristic equation of the homogeneous linear differential equation isThe real general solution is $y=e^{2x}(A\cos 3x+B\sin 3x)$. The first boundary value gives $A=\pi$. At $x=-\pi /2$, $\cos (-3\pi /2)=0$ and $\sin (-3\pi /2)=1$, so $1=e^{-\pi }B$. Hence

Insert the exponential function trial $y=e^{rx}$ into the homogeneous linear differential equation. Its characteristic equation is $r^2+\sqrt{3}r+3=0$, with polynomial rootsTaking real combinations of the two complex exponential solutions produces the general solution

The homogeneous linear differential equation is $y^{\prime \prime }+y=0$. Its characteristic equation is ${\lambda }^2+1=0$, with roots $\pm i$. The general real complementary function is thereforeThe two trigonometric solutions are linearly independent, supplying the two arbitrary constants required for this second-order linear ordinary differential equation.

The characteristic equation of the associated homogeneous linear differential equation is $m^2+7m+12=(m+3)(m+4)=0$. A constant particular integral is $1/2$, soThe first boundary condition gives $A+B=-1/2$. The second gives $A{e}^{-1}+B{e}^{-4/3}=-e^{-1}/2$. Eliminating $A$ yields $B(e^{-4/3}-e^{-1})=0$, hence $B=0$ and $A=-1/2$. Therefore

The equation is not separable, because $x^2-xy=x(x-y)$ cannot be written as a product of a function of $x$ and a function of $y$ on an open region. For example, the ratio $(x^2-x{y}_1)/(x^2-x{y}_2)=(x-y_1)/(x-y_2)$ depends on $x$ for distinct fixed $y_1,y_2$, whereas a separated product would give a constant ratio.

It is not homogeneous in the sense of a homogeneous first-order differential equation: the right-hand side is not a function of $y/x$ alone. It is also not a homogeneous linear differential equation, since moving the $xy$ term to the left leaves a nonzero forcing.

It is linear, since it has the standard linear ordinary differential equation formThus the classifications are (a) no, (b) no, (c) yes.

The associated homogeneous linear differential equation has characteristic roots $2\pm i$, so its complementary function is $e^{2x}(A\cos x+B\sin x)$. For a particular integral, the method of undetermined coefficients suggests $y_p=a{x}^2+bx+c$. Substitution and comparison of coefficients givesso $y_p=25x^2+40x+22$. The general solution is thereforeThe first boundary condition gives $A=-21$. At $x=\pi /2$, the polynomial part already equals the assigned boundary value and the cosine vanishes, so $e^{\pi }B=0$. ThusThis is the unique solution satisfying the two boundary conditions; here the phrase “particular solution” means that selected member of the full family, rather than only the polynomial particular integral.

The characteristic equation of the reduced homogeneous linear differential equation is $m^2-\beta =0$. Its polynomial roots change type with the sign of $\beta$, givingFor $\beta =0$, the zero polynomial root is repeated, so the second independent solution is $t$. For $\beta <0$, the polynomial roots are the complex conjugate pair $\pm i\sqrt{-\beta }$, and the displayed sine and cosine give a real basis of solutions. The constants are arbitrary real numbers. In every case $x(t)=y^{\prime }(t)-\cos t$ reconstructs the corresponding general solution of the coupled system.

The complementary function solves the associated homogeneous linear differential equation. Substituting $y=e^{mx}$ gives the characteristic equationIts distinct polynomial roots giveThe forcing affects the particular solution, which is not needed for the complementary function.

The substitution $y=e^{x}g$ reduces the homogeneous linear differential equation toA first integration gives $g^{\prime }=Cx{e}^{-x}$, and a second gives $g=D-C(x+1)e^{-x}$. HenceThe Wronskian of $1+x,e^x$ is $x{e}^x$, so they are linearly independent on either side of zero. At zero the differential equation has a regular singular point. The displayed solutions extend analytically through it; their linear combinations describe the global solutions when two continuous derivatives are required there. If no matching condition is imposed, the two sides have independent constants.

The homogeneous linear differential equation has solutions $\cos x$ and $\sin x$. Since the forcing is already a homogeneous solution, seek a particular solution proportional to $x\cos x$. Direct differentiation givesThus the general solution of the linear ordinary differential equation isThe first initial condition gives $A=0$. Alsoso the second initial condition gives $B-\frac{1}{2}=-\frac{1}{2}$, hence $B=0$. Therefore

For two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.