"Trivial" differential equations solving, in Hilbert spaces In a book that I'm reading, which is about Hilbert spaces and operators , I've came across some differential equations, and the in the book they simply "guess" the correct results, without explaining it. For example -

Assume that $y(t)$ is an integrable function over $[0,1]$, with the following property:
  $$\lambda y(t) = \int_0^t sy(s) ds +t\int_t^1y(s)ds \quad(\lambda \neq 0 \in \mathbb{C}).$$
  Therefore, we know that $$\lambda y'(t) = ty(t) + \int_t^1y(s)ds-ty(t) = \int_t^1y(s)ds$$ and finally we get the equation: $\lambda y''(t) = -y(t)$. Calculating $t=0$  in the above equations gives us some boundary value conditions: $y'(1)=y(0)=0$. (By multiplying the differential equation by $\bar y$ and integrating from $0$ to $1$, they prove that $\lambda \gt 0$).

Then, in the book they simply say - the solution of the differential equation is
$$y = C_1 \cos\left(\frac{t}{\sqrt{\lambda}}\right)
    + C_2 \sin\left(\frac{t}{\sqrt{\lambda}}\right).$$ I can't understand how and why? How come this result is so trivial? What is the intuition behind it?


Another example, with the following properties: $$\lambda y(t) = t\int_0^ty(s)ds + \int_t^1sy(s)ds,$$ working around the same way above we get that $\lambda y''(t) = y(t)$ with the boundary value conditions - $y'(0) = 0$ and $y(1)=y'(1)$. This time for the case $\lambda \gt 0 $ we get the solution $$y(t) = C_1e^{\frac{t}{\sqrt{\lambda}}} + C_2e^{\frac{-t}{\sqrt{\lambda}}},$$ and for the case $\lambda \lt 0 $, we get a completely different solution: 
  $$y(t) = C_1 \cos\left(t\sqrt{\frac{1}{-\lambda}}\right)
       + C_2 \sin\left(t\sqrt{\frac{1}{-\lambda}}\right).$$

Again, I can't even see how come there are so different solution for the same equation, simply because $\lambda $ has a different sign each time? What is the way to get to the results? Is it simply guessing?
One important thing is, this book is about Hilbert spaces and linear operators only (Ferdholm theory etc..) and does not mention even one theorem from differential equations theory, therefore I assume they did not use any kind of theorem...
 A: It seems to be like you are good with everything up until we get to the mixed-condition differential equation $$\begin{cases} y''=\lambda^{-1} y\\ y'(0)=0,\ y(1)=y'(1).\end{cases}$$ For now, let's forget the conditions and focus just on the ODE. We're looking for a function $y$ whose second derivative is a scaled version of itself. How many functions can you think of that do this? It turns out that all solutions of such an equation are of the form $y(t)=e^{rt},$ where $r$ is to-be-determined. Note that this includes (obviously), exponentials, sines and cosines, and hyperbolic sines and cosines (i.e. the ypes of functions that have this condition). We just need to find $r$. Plugging $e^{rt}$ into the equation, we get $$r^2e^{rt}=\lambda^{-1} e^{rt}.$$ Since $e^t$ is non-vanishing, we can divide through by it, getting the equation $r^2=\lambda^{-1},$ so we have two solutions- $r=\pm1/\sqrt{\lambda}.$ First, let's suppose that $\lambda>0.$ So, we have solutions of the form $e^{t/\sqrt{\lambda}}$ and $e^{-t/\sqrt{\lambda}}.$ Note that we can multiply them both by any number, and they'll still solve the ODE. The same is true  for adding them together. So, the general form of the solution is $$y(t)=c_1e^{t/\sqrt{\lambda}}+c_2e^{-t/\sqrt{\lambda}}.$$ Now, remember that we have two conditions on $y$. If we impose these conditions on our solution, we then find $c_1$ and $c_2.$
Next, suppose that $\lambda<0.$ Then, we have that $r=\pm i/\sqrt{-\lambda}.$ Similar to above, we get that $$y(t)=c_1e^{it/\sqrt{-\lambda}}+c_2e^{-it/\sqrt{-\lambda}}.$$ Recall Euler's formula: $e^{it}=\cos t+i\sin t.$ Applying this to the above, we get, by using the evenness and oddness of $\sin t$ and $\cos t$ (respectively) that
\begin{align*}
y(t)&=c_1\left(\cos\left(\frac{t}{\sqrt{-\lambda}}\right)+i\sin\left(\frac{t}{\sqrt{-\lambda}}\right)\right)\\
&+c_2\left(\cos\left(-\frac{t}{\sqrt{-\lambda}}\right)+i\sin\left(-\frac{t}{\sqrt{-\lambda}}\right)\right)\\
&=c_1\left(\cos\left(-\frac{t}{\sqrt{-\lambda}}\right)-i\sin\left(-\frac{t}{\sqrt{-\lambda}}\right)\right)\\
&+c_2\left(\cos\left(-\frac{t}{\sqrt{-\lambda}}\right)+i\sin\left(-\frac{t}{\sqrt{-\lambda}}\right)\right)\\
&= (c_1+c_2)\cos\left(-\frac{t}{\sqrt{-\lambda}}\right)+(-ic_1+ic_2)\sin\left(-\frac{t}{\sqrt{-\lambda}}\right)\\
&=d_1\cos\left(-\frac{t}{\sqrt{-\lambda}}\right)+d_2\sin\left(-\frac{t}{\sqrt{-\lambda}}\right),
\end{align*}
where $d_1=c_1+c_2$ and $d_2=i(c_2-c_1).$ Once again, we can apply our conditions to find the constants.
