Question

Find two linearly independent solutions to the d.e. y"(x)=0, and then find a solution to y"(x)=1 wiht y(0)=0 and y'(0)=0

Answer 1

Since derivatives lower the degree of polynomials, if you want the second derivative to be zero you have to choose first-degree polynomials.

So, you have

$y''(x)=0 \iff y = c_1x+c_0$

Two polynomials are linearly independent if they are not multiples of each other. So, for example, you might choose $y(x)=1$ and $y(x)=x$ to find two linearly independent solutions.

As for

$y''(x)=1$

we want a second-degree polynomial with leading coefficient 1/2 so that we will get 1 when deriving it twice:

$y''(x)=1 \iff y(x)=\dfrac{x^2}{2}+c_1x+c_0$

If we impose the conditions

$y(0)=0,\quad y'(0)=0$

we have

$y(0)=\dfrac{0^2}{2}+c_1\cdot 0+c_0 = c_0 = 0$

So, our solution will be in this form:

$y''(x)=1,\quad y(0)=0 \iff y(x)=\dfrac{x^2}{2}+c_1x$

To fix $c_1$, we use the second condition:

$y'(x) = x+c_1 \implies y'(0) = c_1 = 0$

So, we have fixed $c_0=c_1=0$ and the solutions is

$\begin{cases}y''(x)=1\\y'(0)=0\\y(0)=0\end{cases} \iff y(x) = \dfrac{x^2}{2}$