$D^4+D^2$
This is in my college made textbook. The solution is $1,x,\cos x,\sin x$ I can understand why cosx and sinx would be part of the solution.
$D^4+D^2 = 0$
$D^2(D^2+1) = 0$
$D= 0, i, -i $
But i don't understand where the 1 and x come from. Anyone else can help?
On
It is important to realize that we are computing the kernel of a linear map $L = D^4 + D^2$ that takes a function $y$ and returns another function $y^{(4)} + y^{(2)} $. As user469000 mentioned, sometimes authors omit the $y$ in these types of questions, but it is useful to remember this is the type of thing being done. Once you get an idea for why we do certain things to solve this type of question, it is quicker to just follow the general procedure. To this end, I will attempt to explain each step of the procedure.
We would like to find linearly independent functions in $\ker D^2(D^2 + 1)$, because this is another way to say $D^2(D^2 + 1)y = 0$. That is, if $y \in \ker D^2(D^2+1)$ then what is $y$? The usual approach is to compute the kernel of each factor $D^2$ and $D^2 + 1$ in the factorization $D^2(D^2+1)$.
First we find linearly independent functions $y$ such that $D^2y = 0$. There are two easy ones that I can think of immediately, those being $1$ and $x$. These are clearly linearly independent and $\dim \ker D^2 = 2$, so we've computed the kernel of $D^2$. There is a theorem (likely in your ODE book) that says the dimension of the kernel of a homogeneous differential equation is its order, which is why we can conclude that $\dim \ker D^2 = 2$. Now the general solution to $D^2 y = 0$ is $y = c_1 \cdot 1 + c_2x$ for constants $c_1, c_2$, hence $\ker D^2 = \text{span }(1, x)$.
Next we find all the linearly independent functions $y$ such that $(D^2 + 1)y = 0$. This is to say $y'' + y = 0 \implies y'' = -y$. If we consider $y = e^{\lambda x}$ for some constant $\lambda$, we see that this equation becomes
\begin{align} \lambda^2 e^{\lambda x} &= -e^{\lambda x}\\ \implies \lambda^2 e^{\lambda x} + e^{\lambda x} &= 0\\ (\lambda^2 + 1)e^{\lambda x} = 0. \end{align} Since the exponential function is nonzero, we must have $\lambda^2 + 1 = 0$. Solutions to this are $i$ and $-i$. When $\lambda = i$, by Euler's formula we have $e^{i x} = \cos x + i \sin x$. Similarly, $e^{-ix} = \cos x - i \sin x$ when $\lambda = -i$. From these two solutions we ultimately get $\cos x$ and $\sin x$ as our linearly independent functions since we can recover these from some linear combination of $e^{ix}$ and $e^{-ix}$, and they are much nicer to work with than the above. Note that we "hide" the imaginary number $i$ in the constants $c_3$ and $c_4$ (referring to user469000's solution here). Similar to before it follows that $\dim \ker (D^2 + 1) = 2$; thus the general solution to $(D^2 + 1)y = 0$ is $y = c_3 \cos x + c_4 \sin x$, meaning $\ker(D^2 + 1) = \text{span }(\cos x, \sin x)$.
We then combine the bases for $D^2$ and $D^2 +1$ to obtain a basis for $D^2(D^2 + 1)$: $\ker D^2(D^2+1) = \text{span } (1, x, \cos x, \sin x)$.
If I've confused you with the above discussion, the important thing to take away from this is that when you have a repeated root ($D^2$ in your question), we usually multiply the previous solution we get by $x$. Thus if we were solving $D^4y = 0$, a basis for its kernel would be $1, x, x^2, x^3$. This is so we can ensure we get all the linearly independent solutions. Otherwise, we are missing out on part of the general solution. Like I said before, once you are comfortable with the algorithm you don't need to worry about all the details above (most people never learn the details regardless).
The question is $(D^4+D^2)y=0$ sometimes in some books author do not mention $y$.
So the auxiliary equation is $M^4+M^2=0$ ,
so $M=0,0,i,-i$
So the C.F. is $(c_1+c_2x)e^{0x}+c_3\cos x+c_4\sin x$ or $c_1+c_2x+c_3\cos x+c_4\sin x$.
Hence the complete solution is $y= c_1+c_2x+c_3\cos x+c_4\sin x$.
Hence the solution set is ${1,x,\cos x,\sin x}.$