0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$L[y]=g(t),\quad y(t_0)=y_0,\quad y'(t_0)=y_0'.$ ----> $(1)$

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 ?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.