so I am revising the Delta Dirac function, and how to solve differential equations with it. I don't want to be employing the Laplace transform at this point, because the question I am practicing specifically says not to use it.
I want to solve, for $a>0$ and $k \in \Bbb{R}$ the initial value problem $$f''(x)-3f'(x)+2f(x)=k\delta(x-a)$$ $f(0)=f'(0)=1$. My method for solving this is generally as follows: split into two regions of the plane, one where $a>0$ and one where $a \le0$. Then, I solve, using the initial boundary conditions for any constants I can, in this case, I can only use the function defined below 0, so can find a value for the constants in that equation. Then after this I ensure continuity of the functions at $a$, so ensure that they equal each other there. After this, I employ the jump condition, so the fact that the $\int_{-\infty}^{\infty}\delta(x-a)dx=1$.
But this is where I get stuck for this situation. I have that $$f_1(x)=Ae^{2x}+Be^x, x>a$$$$f_2(x)=e^x, x\le a$$ But do I then need to be integrating $\int_{-\infty}^{\infty}f''(x)-3f'(x)+2f(x)dx=k$ because this seems really complicated, and to tend towards infinity anyway?
I don't know if I'm integrating the wrong thing here, and that's the issue, but any help is appreciated, thanks.
The generic solutions to the differential equation split on $x<a$ and $x>a$ is $$f(x) = \begin{cases} A e^{2x} + B e^{x} & (x > a) \\ C e^{2x} + D e^{x} & (x < a) \\ \end{cases}$$
Since $a>0,$ applying the continuity conditions $f(0) = f'(0) = 1$ requires that $$\begin{cases} 1 = C + D \\ 1 = 2C + D \\ \end{cases}$$ i.e. $C=0, \ D=1.$ Thus, $$f(x) = \begin{cases} A e^{2x} + B e^{x} & (x > a) \\ e^{x} & (x < a) \\ \end{cases}$$
To get $f''(x)-3f'(x)+2f(x)=k\delta(x-a)$ we want $f$ to be continuous at $a$ but $f'$ to be discontinuous with $f'(a+) - f'(a-) = k.$
The continuity of $f$ at $a$ requires $A e^{2a} + B e^{a} = e^{a}$ i.e. $A e^{a} = 1 - B.$ The step of $f'$ at $a$ requires $2 A e^{2a} + B e^{a} - e^{a} = k$ i.e. $2 A e^{a} = k e^{-a} + (1-B).$ These together give $A = k e^{-2a}, \ B = 1 - k e^{-a}.$
Thus the solution is $$f(x) = \begin{cases} k e^{-2a} e^{2x} + (1 - k e^{-a}) e^{x} & (x > a) \\ e^{x} & (x < a) \\ \end{cases}$$