# Particular solution of a second order differential equation.Tom Apostol's Calculus vol.II section 6.15 exercise 14.

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 $$L(y)=R$$

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

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.