TextBook Fact Check: Transform $\frac{d^2y}{dx^2}+\lambda y=0;\;\;\;y(0)=0,\;y'(1)=0$ into Integral Equation form Want to fact check the solution steps used by my textbook in transforming the above differential equation into an Integral Equation.
Here are the steps:
$$\frac{d^2y}{dx^2}+\lambda y=0; \;\;\; y(0)=0, \; y'(1)=0$$
The solution is $y=\sin \frac{(2n-1)\pi x}{2},\;\;\lambda={[\frac{(2n-1)\pi}{2}]}^2 \;\;\; n=1,2,3...$
Now the step I don't understand is the switch to $\frac{d^2y}{dx^2}=f(x)$
Under the boundary condition specified, they now solved further:
$$y=x\int_1^x f(u)du+\int_x^0uf(u)du$$
The Integral formulation becomes:
$$y(x)=\lambda\int_0^1K(x, u)y(u)du$$
Where:
$K(x, u)=x \;\;\;\;\; 0\leq x\leq u \;\;\;\;\; 1\leq u\leq x$
$K(x, u)=u \;\;\;\;\; u\leq x\leq1 \;\;\;\;\; x\leq u\leq0$
Where did they get it wrong. And what's the right step to take?
 A: Let us very slightly modify the expression you give:
$$\tag{1}y(x)=x g(x) -\int_0^xuf(u)du \ \ \ \ \text{with} \ \ \ \ g(x):=\int_1^x f(u)du$$
Let us differentiate it with respect to $x$:
$$\tag{2}y'(x)=g(x) + xg'(x) - xf(x)$$
Because $g'(x):=f(x)$, relationship (2) boils down to: 
$$\tag{3}y'(x)=g(x)$$
Differentiating once more: 
$$\tag{4}y''(x)=g'(x)=f(x),$$
what was desired.
Moreover, we retrieve the initial conditions $y(1)=y'(1)=0$. Here is how: 


*

*Making $x=0$ in (1) gives: $y(1)= 0g(0) -\int_0^0uf(u)du=0.$

*Making $x=0$ in (3) gives $y'(1)=g(1)=\int_1^1f(u)du=0.$ 
I have not commented the rest. Do you agree with the transformation into the classical "kernel" $K$ formulation? (Caution : your last column has to be removed). A remark: $K$ can be written
$$K(x,t)=min(x,t).$$
Its relationship with the second derivative is well known: see my answer in (https://math.stackexchange.com/q/1897846).
A question: I haven't understood what you call "wrong' in all this formulation.
A: I have finally gotten it with the help of @JeanMarie
My confusion was the switch from $$\frac{d^2y}{dx^2}+\lambda y=0$$ to $$\frac{d^2y}{dx^2}=f(x) $$
I've finally come to realise that $$f(x)$$ was equated to $$-\lambda y(x) $$
i.e. $f(x)=-\lambda y(x) $
The Integral form was then calculated for the equation $ \frac{d^2y}{dx^2}=f(x)$
$$y(x)=\int_1^0K(x, u)f(u)du\;\;\;\;\;\; (1)$$
Note however that $f(x)=-\lambda y(x) $
Thus $$f(u)$$ transform into into $$-\lambda y(u) $$
And equation (1) becomes:
$$y(x)=\lambda\int_0^1K(x, u)y(u)du\;\;\;\;\;\; (2)$$
Where:
$K(x, u)=x \;\;\;\;\; 0\leq x\leq u \;\;\;\;\; 1\leq u\leq x$
$K(x, u)=u \;\;\;\;\; u\leq x\leq1 \;\;\;\;\; x\leq u\leq0$
