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Where $y_1$ is nonzero, the Wronskian identity $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ becomes the first-order equationIts integrating factor is $1/y_1$, giving $(y_2/y_1{)}^{\prime }=W/y_1^2$. Integration proves the reduction of order formulaTaking $C_0=0$ gives the displayed representative in the question, with $y_2(x_0)=0$ when $y_1(x_0)\ne 0$. Changing the lower limit adds a multiple of $y_1$ and does not change the linearly independent solution modulo the known one. The scale of $W$ separately fixes the normalization of $y_2$.

Differentiate the determinant and use the original second-order equation for each solution:This proves Abel's identity $W^{\prime }+pW=0$ rather than merely quoting it.

For the specific equation, substituting $y_1=1-x$, $y_1^{\prime }=-1$, $y_1^{\prime \prime }=0$ leaves $(1-x^2)-(1+x)(1-x)=0$. Away from zero its normalized first-derivative coefficient is $p(x)=x-1/x$, soAlthough the equation is singular at zero, this expression extends analytically there. Near zero, $y_1$ is nonzero and the reduction of order integrand is regular. With the prescribed lower limit,The integrand through degree three isAfter integrating and multiplying by $1-x$,The condition $y_2^{\prime \prime }(0)=1$ fixes $C=1$, and the lower limit already gives $y_2(0)=0$. Therefore the first three nonzero terms areThe representation genuinely solves the second-order equation: writing $L[y]=y^{\prime \prime }+p{y}^{\prime }+qy$, differentiating its Wronskian gives $W^{\prime }+pW=y_{1}L[y_2]-y_{2}L[y_1]=y_{1}L[y_2]$. Its left-hand side is zero and $y_1\ne 0$ near zero. Analyticity then extends the original unnormalized equation to zero itself. Its Wronskian with $y_1$ is nonzero for sufficiently small $x\ne 0$, proving linear independence. The vanishing Wronskian at the singular point zero does not contradict linear independence on a regular interval. The integral representation is used locally near zero, before the zero of $y_1$ at one.

Two functions are linearly dependent if constants $c_1,c_2$, not both zero, satisfy $c_1{y}_1+c_2{y}_2=0$ identically. Differentiating also gives $c_1{\dot{y}}_1+c_2{\dot{y}}_2=0$. Since the determinant of these two equations is the nonzero Wronskian $W$, both constants must vanish. Hence the two solutions are linearly independent.

Differentiating the Wronskian and substituting the homogeneous linear differential equation givesIntegrating proves Abel's identity:For variation of parameters, impose ${\dot{a}}_1{y}_1+{\dot{a}}_2{y}_2=0$ on the proposed particular solution. Then $\dot{y}=a_1{\dot{y}}_1+a_2{\dot{y}}_2$, and substitution into the inhomogeneous differential equation leaves ${\dot{a}}_1{\dot{y}}_1+{\dot{a}}_2{\dot{y}}_2=f$. Solving this two-by-two linear system givesArbitrary constants of integration merely add a homogeneous solution.

For the constant positive coefficient $q={\omega }^2$, take $\omega >0$. Direct differentiation verifies that $y_1=\cos \omega t$ and $y_2=\sin \omega t$ solve the homogeneous linear differential equation; their Wronskian is $W=\omega$. The requested integrands areTaking $t_0=0$ yieldsThus $|a_1(t)|\sim t/(2\omega )$ grows with power one, whereas $a_2$ is bounded. Their combination simplifies to $y_p=-t\cos (\omega t)/(2\omega )+\sin (\omega t)/(2{\omega }^2)$; the final sine term is itself a homogeneous solution. The secular term is the response to resonance in a differential equation.