# Finding the properties of solutions of a second order non homogeneous ODE

Show that the solution of the initial value problem $$L[y] = y'' + p(t)y' + q(t)y = g(t), y(t_0) = y_0, y'(t_0) = y'_0$$ can be written as $y = u(t) + v(t)$, where $u$ and $v$ are solutions of the two initial value problems $$L[u] = 0, u(t_0) = y_0, u'(t_0) = y'_0$$ $$L[v] = g(t), v(t_0) = 0, v'(t_0) = 0$$ respectively. In other words, the nonhomogeneities in the differential equation and in the initial conditions can be dealt with separately.

What i tried

Since $L[u] = 0$, $u(t)$ is the complemntary solution, hence is the solution to the homogeneous eqn $$L[u] = u'' + p(t)u' + q(t)u = 0, u(t_0) = u_0, u'(t_0) = u'_0$$ and $v(t)$ is the particular solution.

So to show that $u(t)$ is the complementary solution, i subusituted $u(t)$ to the above ODE, Im also thinking should i solve the equation$$y'' + p(t)y' + q(t)y = 0, y(t_0) = y_0, y'(t_0) = y'_0$$ but im unsure how to go about doing it. Could anyone explain. Thanks

$L[y]=g(t),\quad y(t_0)=y_0,\quad y'(t_0)=y_0'.$ ----> $(1)$
Let $y_c(t)$ be the solution of $L[y]=0,y(t_0)=y_0,\quad y'(t_0)=y_0'$ and $y_p(t)$ be the particular solution. Then
$L[y_c+y_p]=L[y_c]+L[y_p]=0+g(t)=g(t)$ ($\because$ L is a linear operator),
and $(y_c+y_p)(t_0)=y_c(t_0)+y_p(t_0)=y_0+0=y_0$. Similarly the other condition. Hence $y_c+y_p$ is the solution of $(1)$. Does this help you ?
• you basically check whether $y_c+y_p$ satisfies the non-homogeneous problem or not. – Hirak Feb 21 '15 at 9:48
• the operator $L$ is not self-adjoint and is not needed fro the principle of superposition to hold; only linearity of $L$ and of the boundary conditions are used. – abel Feb 21 '15 at 12:29