Find general solution of the differential equation (linear algebra) How to find the general solution of a differential equation?
$$8y''' + 8y' = 0$$
I literally know nothing about solving differential equations. Honestly I have never been taught about differential equations and I don't know why my professor threw this question into the homework assignmnent.
 A: Here's how a person might be able to solve this, even without ever being taught anything about differential equations: 
First, such a person might know that constant functions have derivative zero, and might then realize that if $y$ is constant, then $y'$ and $y'''$ are both identically zero, so $8y'''+8y'=0$. Thus, every constant function is a solution. 
Second, such a person may know that the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. It follows immediately that the second derivative of $\sin x$ is $-\sin x$, so, if $y=\sin x$, then $y''=-y$, whence $y''+y=0$, whence $8y''+8y=0$. Now differentiating gives $8y'''+8y'=0$, so $y=\sin x$ is a solution. 
But then the same argument applies to $B\sin x$ for any constant $B$, so $y=B\sin x$ is a solution. And also, the second derivative of $\cos x$ is $-\cos x$, so, for any constant $C$, $y=C\cos x$ is a solution. 
Then, knowing that $(f+g)'=f'+g'$, our hypothetical person may come to realize that $$y=A+B\sin x+C\cos x$$ is a solution for all constants $A,B,C$. 
This is where I get stuck – I'm not sure how the person who has never been taught anything about solving differential equations could conclude that every solution is of this form, so that we have found the general solution. 
A: Hint:
This differential equation is equivalent to $y'''+y'=0$, and by making the substitution $u=y'$, we find that it can be expressed as $$u''+u=0$$ Does that look easier to solve?
