# Another form of Second Order ODE solution with two distinct real roots

Suppose i have ODE:

$$y''-4y=0$$

We know that the solution is

$$y=c_1 e^{2x} + c_2 e^{-2x}$$

And I realize that this also has the following solution

$$y=c_1 \cosh(2x) + c_2 \sinh(2x)$$

are the two solutions equivalent?

So, if i have this ODE

$$ay''-by=0\qquad a>0, b>0$$

May I write the general equation as follows?

$$y=c_1 \cosh(\sqrt{\frac{b}{a}}x) + c_2 \sinh(\sqrt{\frac{b}{a}}x)$$

Thanks.

• Yes, the two solutions are equivalent (with different values of $c_1$ and $c_2$). If you have the first solution, you can always re-write it in the form of the second solution, and vice versa. Yes, your last expression is the general solution. – John Barber May 11 at 22:08
• It's simply a changing of the basis. Instead of $(e^{2x},e^{-2x})$ you take $(\cosh(2x),\sinh(2x))$. Of course coordinates, $c_1,c_2$, will change. – Wang May 11 at 22:11

Any linear combination of $$e^{\lambda_1}$$ and $$e^{\lambda _2}$$ is a solution