About solution of equation $y''-C^{2}y=f(x)$

Question

I need some explanation of this result

Problem :

$$\begin{equation} \begin {cases} y''(x)-C^{2}y(x)=f(x)~~~~,x\in ]-1,0[\\ y(0)=a\\ y(-1)=b \\ y'(0)=\lambda \end{cases} \end{equation}$$

And the results is :

$$y(x)=\alpha e^{-xC}+\beta e^{(x+1)C}+v(f)(x)$$

Where

$v(f)(x)=\frac {1}{2}\displaystyle\int\limits_{-1}^{x}e^{(x-t)B}\dfrac{f(t)}{C}dt+\frac {1}{2}\displaystyle\int\limits_{x}^{0}e^{(t-x)B}\dfrac{f(t)}{C}dt%)$

$\textbf{My attempts : }$

First we will solve homogeneous equation : $$y''-C^{2}y=0$$ This implies $y_{h}=\alpha e^{-xC}+\beta e^{xC}$

Now let find particular solution (using wronski $W(x)$).

Such that $\{ e^{-xC}, e^{xC}\}$ are base of solution then : $$y_{p}=y_{1}e^{-xC}+y_{2} e^{xC}$$

Where :

$y_{1}(x)=-\int\dfrac{e^{xC}f(x)}{W}dx$

And

$y_{1}(x)=\int\dfrac{e^{-xC}f(x)}{W}dx$

And

$$W= \begin{vmatrix} e^{-xC} & e^{xC}\\ (e^{-xC})^{\prime} & (e^{xC})^{\prime} \end{vmatrix} =2C $$

So :

$y_{1}(x)=-\int\limits_{-1}^{x}\dfrac{e^{tC}f(t)}{2C}dt$

$y_{1}(x)=\int\limits_{-1}^{x}\dfrac{e^{-tC}f(t)}{2C}dt$

And the solution become :

$$y(x)=\alpha e^{-xC}+\beta e^{(x+1)C}-e^{-xC}\int\limits_{-1}^{x}\dfrac{e^{tC}f(t)}{2C}dt+e^{xC}\int\limits_{-1}^{x}\dfrac{e^{-tC}f(t)}{2C}dt = \alpha e^{-xC}+\beta e^{(x+1)C}-\int\limits_{-1}^{x}\dfrac{e^{(t-x)C}f(t)}{2C}dt+\int\limits_{-1}^{x}\dfrac{e^{(x-t)C}f(t)}{2C}dt$$

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Where is my mistake ? Looked to the bound of integration?

Answer 1

HINT

Here it is another possible approach:

\begin{align*} y'' - C^{2}y = f(x) & \Longleftrightarrow (y'' - Cy') + (Cy' - C^{2}y) = f(x)\\\\ & \Longleftrightarrow (y' - Cy)' + C(y' - Cy) = f(x)\\\\ & \Longleftrightarrow w' + Cw = f(x)\\\\ & \Longleftrightarrow (e^{Cx}w)' = e^{Cx}f(x)\\\\ & \Longleftrightarrow e^{Cx}w = \int e^{Cx}f(x)\mathrm{d}x + A\\ & \Longleftrightarrow y' - Cy = e^{-Cx}\int e^{Cx}f(x)\mathrm{d}x + Ae^{-Cx} \end{align*}

Can you take it from here?