Respuesta :
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
[tex]y''(x)=0 \iff y = c_1x+c_0[/tex]
Two polynomials are linearly independent if they are not multiples of each other. So, for example, you might choose [tex]y(x)=1[/tex] and [tex]y(x)=x[/tex] to find two linearly independent solutions.
As for
[tex]y''(x)=1[/tex]
we want a second-degree polynomial with leading coefficient 1/2 so that we will get 1 when deriving it twice:
[tex]y''(x)=1 \iff y(x)=\dfrac{x^2}{2}+c_1x+c_0[/tex]
If we impose the conditions
[tex]y(0)=0,\quad y'(0)=0[/tex]
we have
[tex]y(0)=\dfrac{0^2}{2}+c_1\cdot 0+c_0 = c_0 = 0[/tex]
So, our solution will be in this form:
[tex]y''(x)=1,\quad y(0)=0 \iff y(x)=\dfrac{x^2}{2}+c_1x[/tex]
To fix [tex]c_1[/tex], we use the second condition:
[tex]y'(x) = x+c_1 \implies y'(0) = c_1 = 0[/tex]
So, we have fixed [tex]c_0=c_1=0[/tex] and the solutions is
[tex]\begin{cases}y''(x)=1\\y'(0)=0\\y(0)=0\end{cases} \iff y(x) = \dfrac{x^2}{2}[/tex]
