# Understanding Hyperbolic / Exponential General Solutions for Linear Homogeneous ODEs

In my PDE class, we are learning to solve linear homogeneous PDEs by separating variables and solving(?) for our ODEs. Working with a case with $\lambda > 0$ we are able to come to the following general solutions.

$\frac{d^2h}{dy^2}=-\lambda h \implies h=c_1 \cos{\sqrt{\lambda}y}+c_2 \sin{\sqrt{\lambda}y}$

$\frac{d^2h}{dy^2}=\lambda h \implies h=c_1 \cosh{\sqrt{\lambda}y}+c_2 \sinh{\sqrt{\lambda}y} = ce^{\sqrt{\lambda}y}$

My question is why do we assume these general solutions based off our ODE's? When does it make more sense to use the hyperbolic general solution for $\frac{d^2h}{dy^2}=\lambda h$ vs the much simpler exponential solution? Also if you could include a derivation for how we got those general solutions from the ODE's that would be fantastic.

$c_1\cosh{\sqrt{\lambda}y}+c_2 \sinh{\sqrt{\lambda}y} \neq ce^{\sqrt{\lambda}y}$

in fact :

\begin{align} c_1\cosh{\sqrt{\lambda}y}+c_2 \sinh{\sqrt{\lambda}y} ={c_1 \over 2}(e^{\sqrt{\lambda}y} + e^{-\sqrt{\lambda}y}) + {c_2 \over 2}(e^{\sqrt{\lambda}y} - e^{-\sqrt{\lambda}y}) = k_1 e^{\sqrt{\lambda}y} + k_2 e^{-\sqrt{\lambda}y}\\ \end{align}

edit : in general a linear second order ODE with constant coefficients is of the form $y'' + ay' + by + c = 0$

now something very important to keep in mind is that the set of solutions of this kind of equation is a vector space of dimension $2$ meaning if you find $2$ independent solutions you're done. because two independent solutions form a basis in a two-dimensional vector space. meaning : you can write any solution as a linear combination of them.

now how to find $2$ independant solutions ?

assume a solution is of the form $y = e^{rx}$

after plugging this in the original equation and dividing by $e^{rx}$ you get a quadratic equation which is easy to solve.

you usually either find $2$ different roots or a double root.

if you find $2$ different real roots then it's trivial that $e^{r_1x}$ and $e^{r_2x}$ are independent

if you find $2$ complex roots then one has got to be the conjugate of the other and here another very important thing you must keep in mind as well : is that if a complex function is a solution to the above equation then its real and imaginary parts are all solutions. and this is easily verifiable. and that's why in this case $\cos$ and $\sin$ appear in the solution.

there is a third case where you get only one root but you could still find a secod solution by a method called 'variation of parameters'