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I have this ODE : $y^{''}(x)-Ay(x)=Bx \delta_{0}(x)$ where $A,B$ are constants and $\begin{equation} \delta_{0}= \begin{cases} \infty & \text{if $x=0$}, \\ 0& \text{else}. \end{cases} \end{equation}$ Particular solution by variation of constants involves $\int Bx e^{\sqrt{A}x}\delta_{0}(x)\,dx $ , which is equal to zero and I can not achieve any particular solution using this method. How can I find a particular solution for this case?

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"In the beginning, God created the heavens and $\delta_0=\begin{cases} \infty & \text{if $x=0$} \\ 0 & \text{else} \end{cases}$.${}\ {}$" ${}\quad{}$ "In the beginning, God created the heavens and $\delta_0=\begin{cases} \infty & \text{if $x=0$}, \\ 0 & \text{else}. \end{cases}$ ${}\ {}$" ${}\quad{}$ I think the second format here is preferable, and I edited the question accordingly. –  Michael Hardy Nov 25 '12 at 20:07
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Where did this ODE come from? I wonder if it's possible to make sense of something like $Bx\delta_0(x)$ other than by saying it's $0$. –  Michael Hardy Nov 25 '12 at 20:13
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Note that $\int f(x)\delta_0(x)~dx=f(0)u(x)+C$ , where $u(x)$ is an unit-step function. –  doraemonpaul Nov 25 '12 at 21:18

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I agree that there is no other way to make sense of something like $x\delta(x)$ other than setting it to $0$. You can be a bit more explicit if you proceed solving the ODE in the following way. You can solve it in the negative and positive parts of the real line and then try gluing it together at $x=0$ by appealing to continuity of $y$ as well as deriving a jump condition for $y'$ by integrating the ODE.

For $x<0$, the solution to the ODE is of the form $$C_1 \cos(Ax) + C_2 \sin(Ax),$$ similarly for $x>0$ it looks like $$C_3 \cos(Ax) + C_4 \sin(Ax).$$ Now $y$ has to be continuous at $y=0$ since otherwise our ODE would involve a derivative of the delta-function (some hand-waving involved here but I never know how to make arguments involving the $\delta$-function precise). This yields $C_2 = C_4$. To get the relation between $C_3$ and $C_1$ integrate the ODE from $-\epsilon$ to $\epsilon$, use the continuity of $y$ and then take the limit as $\epsilon$ goes to zero. This would yeild that $y'$ is also continuous at $0$ which would then mean that $C_1 = C_3$ and thus a $0$ particular solution.

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In fact the particular solution of this inhomogeneous linear ODE should be equal to $0$ as $x\delta_0(x)=0$ . http://en.wikipedia.org/wiki/Dirac_delta_function#Algebraic_properties has been verified this.

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