I was trying to convert this Boundary Value problem to an integral equation(Fredholm Integral Equation)
$y'' + y' + y = 0 ,y(0) = 0, y(1) = 1 \rightarrow (*)$
I tried but got stuck,here Integrating the above ODE we get
$\int_{0}^{x}y''(x) dx + \int_{0}^{x}y'(x) dx + \int_{0}^{x}y dx = c$
$y'(x) - y'(0)+y(x)- y(0) + \int_{0}^{x}y dx = c$
at $x = 1$
$y'(1) - y'(0) + y(1) - y(0) + \int_{0}^{1}y dx = c$
So $c = y'(1) - y'(0) + 1 + \int_{0}^{1}y dx$
Next substituting this value of $c$ in $(*)$,we get
$y(x) = -\int_{0}^{x}y dx - \int_{0}^{x}(x-t)y(t)dt + xy'(1) + x + x\int_{0}^{1}y dx$
I think I am making mistake somewhere as the form is
$y = \int_{0}^{1} K(x,t)y(t)dt$,
any help!
Consider a function $g\in C^2([0,1])$ with $g'(0)=0$ and $g'(x)>0$ for $x>0$ (for example $g(x)=x^2$). Multiply the equation by $g$ and integrate. You get $$0=\int_0^x(y^{\prime\prime}g+y^{\prime}g+yg)\, dt =y^{\prime}(x)g'(x)-\int_0^xy^{\prime}g'\, dt-y(x)g(x)\\+\int_0^x yg'\, dt+\int_0^xyg\, dt =y^{\prime}(x)g'(x)-y(x)g'(x)\\+\int_0^xyg^{\prime\prime}\, dt-y(x)g(x)+\int_0^x yg'\, dt+\int_0^xyg\, dt.$$ Hence, $$y'(x)=\frac1{g'(x)}\left(-y(x)g'(x)+\int_0^xyg^{\prime\prime}\, dt-y(x)g(x)+\int_0^x yg'\, dt+\int_0^xyg\, dt\right).$$ Integrate and use Fubini's theorem (and relabel variables) $$y(x)=\int_0^x\frac1{g'(t)}\left(-y(t)g'(t)+\int_0^tyg^{\prime\prime}\, ds-y(t)g(t)+\int_0^t yg'\, ds+\int_0^tyg\, ds\right)dt \\=\int_0^x y(t)\left(-1-\frac{g(t)}{g'(t)}+(g^{\prime\prime}(t)+g'(t)+g(t))\int_t^x \frac1{g'(s)}\, ds\right)dt,$$ where we used the fact that $0\le s\le t\le x$.