$y'' - y = 0 $ for $0This is an odd numbered problem from my textbook, so I have the solution I'm just curious as to how one gets the answer?
Can someone walk me through it?
Thanks :)
 A: For our characteristic equation, we have:
$$m^2 - 1 = 0 \rightarrow m_1 = -1, m_2 = 1$$
This means our solution is:
$$y = c_1e^{-x} + c_2 e^x$$
Plug in the initial conditions to solve for $c_1$ and $c_2$.


*

*$y(0) = c_1 + c_2 = 1$

*$y(1) = c_1e^{-1} + c_2e = -4$


Upon solving these, we arrive at:


*

*$c_1 = \dfrac{4e +1}{e^2-1} +1$

*$c_2 = -\dfrac{4e+1}{e^2-1}$


You should be able to handle it from here.
We can look at the solution curve as:

We can also look at the phase portrait as:

A: This is a standard CCLDE, or constant coefficient linear differential equation, so we use the characteristic equation to solve it:
Let $y'' - y = 0$. The characteristic polynomial is:
$\lambda^{2} - 1 = 0$
So our roots are:
$\lambda = \pm 1$
Which means our fundamental set of solutions is:
$\{e^{1x}, e^{-1x}\} = \{e^{x}, e^{-x}\}$
Hence, our general solution takes the form:
$y = C_{1}e^{x} + C_{2}e^{-x}$
Plugging into the equation with known values to determine $C_{1}$ and $C_{2}$, we find:
$y(0) = C_{1} + C_{2} = 1 \implies C_{1} = 1-C_{2}$
$y(1) = C_{1}e + C_{2}e^{-1} = (1-C_{2})e + C_{2}e^{-1} = -4$
So
$\frac{e^{2}-1}{e}(C_{2}) = 4 + e$
Can you finish solving for $C_{1}$ and $C_{2}$ from here?
