If $L(y) = y”+ ay’ + by $ the place a and b are constants, let f be the prticulr answer of $L(y)=0$ satisfying the circumstances $f(0) = 0 $ and $f'(0) = 1$.Present {that a} explicit answer off $L(y) = R $ is given by the method $$ y_1(x) = int_c^x f(x-t)R(t)dt $$ for any selection of c.

Query : One can present this by substituting $y$ with $ y_1 $ in L(y) and utilizing Leibniz rule for integration and $f(0) = 0 $ and $f'(0) = 1$. I’m questioning if one can clear up this utilizing theorem (Tom Apostol’s Calculus vol.II Theorem 6.11.)

THEOREM 6.11 : Let $u_1,…….u_n$ be n unbiased options of the homogenous nth order linear differential equation $L(y) = 0$ on n interval J. Then a selected answer y_1 of the nonhomogenous equation $L(y) = R$ is given by the method $$ y_1(x) = sum_{ok=1}^n u_k (x) v_k (x),$$the place $v_1,…..v_n $ are the entries of the n x 1 column matrixx v decided by the equation $$v(x) = int_c^x R(t)W(t)^{-1}start{pmatrix} vdots 1 finish{pmatrix} dt $$the place W is the Wronskian matrix of $u_1,…..,u_n$ and c is any level in J.

Mixed with Abel’s method $detW(x) = detW(c)exp[int_c^x P_1(t) dt]$ $cepsilon J$.

My considering is that since f is an answer of L(y) then f has the shape $f(x) = c_1e^{-ax/2}u_1(x) + c_2e^{-ax/2}u_2(x)$.Taking the Wronskian matrix of $v_1 = e^{-ax/2}u_1(x)$ and $v_2 = e^{-ax/2}u_2(x)$ and $detW(0) = -u_1(0)e^{-ax}/c_1$(utilizing the truth that $f(0) = 0 $ and $f'(0) = 1$.) in Abel’s method we then have a selected answer $ y_1(x) = sum_{ok=1}^2 g_k (x) v_k (x),$ the place $g_1$ and $g_2$ are the entries of the two x 1 column matrix given by $$g(x) = int_c^x R(t)(-c_1e^{at}/u_1(0))start{pmatrix}-e^{-at/2}u_2(t) e^{-at/2}u_1(t) finish{pmatrix} dt $$.

My query is then, if we will proceed from that considering and present that $$ y_1(x) = int_c^x f(x-t)R(t)dt $$ is a selected answer of $L(y)=R$