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If $L(y) = y''+ ay' + by $ where $a$ and $b$ are constants, let $f$ be the particular solution of $L(y)=0$ satisfying the conditions $f(0) = 0 $ and $f'(0) = 1$. Show that a particular solution of $L(y) = R $ is given by the formula $$ y_1(x) = \int_c^x f(x-t)R(t)dt $$ for any choice of $c$.

Question : One can show this by substituting $y$ with $ y_1 $ in $L(y)$ and using Leibniz rule for integration and $f(0) = 0 $ and $f'(0) = 1$. I am wondering if one can solve this using theorem (Tom Apostol's Calculus vol.II Theorem 6.11.)

THEOREM 6.11 : Let $u_1,.......u_n$ be $n$ independent solutions of the homogeneous $n$th order linear differential equation $L(y) = 0$ on an interval $J$. Then a particular solution $y_1$ of the non-homogeneous equation $L(y) = R$ is given by the formula $$ y_1(x) = \sum_{k=1}^n u_k (x) v_k (x),$$ where $v_1,.....v_n $ are the entries of the $n \times 1$ column matrix $v$ determined by the equation $$v(x) = \int_c^x R(t)W(t)^{-1}\begin{pmatrix}0\\ \vdots \\0\\1\end{pmatrix} dt $$ where $W$ is the Wronskian matrix of $u_1,.....,u_n$ and $c$ is any point in $J$.

Combined with Abel's formula $\det W(x) = \det W(c)\exp[\int_c^x P_1(t) dt]$ $c\in J$.

My thinking is that since $f$ is a solution of $L(y)$ then $f$ has the form $f(x) = c_1e^{-ax/2}u_1(x) + c_2e^{-ax/2}u_2(x)$. Taking the Wronskian matrix of $v_1 = e^{-ax/2}u_1(x)$ and $v_2 = e^{-ax/2}u_2(x)$ and $detW(0) = -u_1(0)e^{-ax}/c_1$ (using the fact that $f(0) = 0 $ and $f'(0) = 1$.) in Abel's formula we then have a particular solution $ y_1(x) = \sum_{k=1}^2 g_k (x) v_k (x),$ where $g_1$ and $g_2$ are the entries of the 2 x 1 column matrix given by $$g(x) = \int_c^x R(t)(-c_1e^{at}/u_1(0))\begin{pmatrix}-e^{-at/2}u_2(t) \\e^{-at/2}u_1(t) \end{pmatrix} dt .$$

My question is then, if we can continue from that thinking and show that $$ y_1(x) = \int_c^x f(x-t)R(t)dt $$ is a particular solution of $L(y)=R$

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    $\begingroup$ Since $y_1$ is given, why don't you simply compute $L(y_1)$ ? $\endgroup$
    – Surb
    Apr 1, 2020 at 7:44
  • $\begingroup$ I was wondering if there is another way, without using Leibniz integral rule! $\endgroup$ Apr 1, 2020 at 7:47
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    $\begingroup$ Leibniz rule ? there is no Leibniz rule here. $\endgroup$
    – Surb
    Apr 1, 2020 at 7:48
  • $\begingroup$ Dont we need it to calculate the derivative of the integral?because of f(x-t). $\endgroup$ Apr 1, 2020 at 7:52
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    $\begingroup$ $$y'_1(x)=\frac{\mathrm d}{\mathrm d x}\int_0^x f(x-t)R(t)\,\mathrm d t=f(0)R(0)+\int_0^x f'(x-t)R(t)\,\mathrm d t$$$$=\int_0^x f'(x-t)R(t)\,\mathrm d t$$ $\endgroup$
    – Surb
    Apr 1, 2020 at 8:24

1 Answer 1

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This task is related to the construction that if $$ L(y)=y^{(n)}+a_{n-1}y^{(n-1)}+...+a_1y+a_0 $$ with $a_k=const.$, and $f$ is a solution of $L(y)=0$ with $y(0)=y'(0)=...=y^{(n-2)}(0)=0$ and $y^{(n-1)}=1$, and $\theta$ is the Heaviside unit jump function, then the product $y_1=\theta f$ solves $$ L(\theta f)=\delta $$ with the Dirac delta on the right side. Any other inhomogeneous problem $L(y)=R$ has then a solution $y=(\theta f)\ast R$, so that $$ y(x)=\int_{\Bbb R}(\theta f)(x-t)R(t)dt=\int_{-\infty}^xf(x-t)R(t)dt $$ One would check what happens if the integration interval is cut at some $c$ at the lower end, presumably $$ y_c(x)=\int_{-\infty}^cf(x-t)R(t)dt $$ would have to be a homogeneous (or complementary) solution.

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