Find the expression for function $g(x)$ I was given that $g(0)=1, g'(0)=0$ and $g''(x)-g(x)=0$. Will I be able to get the exact expression for $g(x)$? If so, how can I do that? 
 A: For such a kind of equation where is a linear combination of higher derivatives a function, generally you expect an exponential function, or trigonometric functions (sine and cosine) to be solution. 
So in that case it is assumed that $e^{mx}$ is a particular solution and substituted in place of g(x). So you get $$(m^2 - 1)e^{mx} = 0$$, so you get $(m^2 - 1) = 0$ which is called the auxiliary equation.
$m = -1 \mbox{ or } +1$ solves it as $m^2 - 1 = (m-1)(m+1) = 0$.
Now the general solution is a linear combination of these two functions, i.e.
$$ g(x) = c_1 e^x + c_2 e^{-x}$$, where $c_1$ and $c_2$ are arbitrary constants. If you now apply the initial conditions i.e $g(0) = 1$ and $g'(0) = 0$ you get two equations namely : 
$$ c_1 + c_2 = 1$$ and $$c_1 - c_2 = 0$$, if you solve these real constants (hint add them and subtract these equations), then you get $$c_1 = c_2 =\frac{1}{2}$$. So the solution $$g(x) =  \frac{1}{2} e^x + \frac{1}{2} e^{-x}.$$
Verify that it satisfies every condition you have for $g$. 
A: The standard way of solving equations like this (equations of $g'' + a g' + b g = 0$) is to set $g(x) = e^{\lambda x}$ and try to find the appropriate $\lambda$. The equation you get is $$\lambda^2 + a\lambda + b = 0$$
which you can solve for $\lambda$. Then you have $3$ options:


*

*If there exist $2$ distinct $\lambda_1\ne \lambda_2\in\mathbb R$ that solve the equation, then any solution to the original equation will take the form
$$g(x) = Ae^{\lambda_1 x}+Be^{\lambda_2x}$$

*If there exists one real $\lambda$ that solves the equation, then any solution to the original equation will take the form $$g(x) = (Ax+B)e^{\lambda x}$$

*If the solutions to $\lambda^2 + a\lambda + b = 0$ are imaginary, then they are of the form $\lambda_{1,2} = a\pm bi$. In that case, any solution to the original equation will take the form $$g(x) = e^{a}\left(A\cos(bx) + B\sin(bx)\right)$$


In all three cases, you know that $g(x)$ is a function, depending on $2$ real numbers $A$ and $B$. Plugging in the known values of $g(0)$ and $g'(0)$ helps you determine what $A$ and $B$ are.
