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 ?

  • $\begingroup$ So basically u find the sum of the Complementary and particular solutions to see whether it matches the solution of the original non homogenous ODE? $\endgroup$ – ys wong Feb 21 '15 at 9:45
  • $\begingroup$ you basically check whether $y_c+y_p$ satisfies the non-homogeneous problem or not. $\endgroup$ – Hirak Feb 21 '15 at 9:48
  • $\begingroup$ 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. $\endgroup$ – abel Feb 21 '15 at 12:29
  • $\begingroup$ abel: could u please help me with this problem as well. Thanks math.stackexchange.com/questions/1164855/… $\endgroup$ – ys wong Feb 25 '15 at 13:41

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