# How do you find all solutions of the differential equation (d^2y)/(dx^2)=3y?

##### 1 Answer
Dec 31, 2016

$\setminus \setminus \setminus \setminus \setminus y = A {e}^{\sqrt{3} x} + B {e}^{- \sqrt{3} x}$

Where $A , B$ are arbitrary constants.

#### Explanation:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 3 y \implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + 0 \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y = 0$

This is a Second Order Homogeneous Differential Equation which we solve as follows:

We look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

${m}^{2} + 0 m - 3 = 0$
$\therefore {m}^{2} - 3 = 0$
$\therefore m = \pm \sqrt{3}$

Because this has two distinct real solutions $\sqrt{3}$ and $- \sqrt{3}$, the solution to the DE is;

$\setminus \setminus \setminus \setminus \setminus y = A {e}^{\sqrt{3} x} + B {e}^{- \sqrt{3} x}$

Where $A , B$ are arbitrary constants.

-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Verification:

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus y = A {e}^{\sqrt{3} x} + B {e}^{- \sqrt{3} x}$

$\therefore \setminus y ' = A \sqrt{3} {e}^{\sqrt{3} x} - B \sqrt{3} {e}^{- \sqrt{3} x}$

$\therefore y ' ' = A \sqrt{3} \sqrt{3} {e}^{\sqrt{3} x} + B \sqrt{3} \sqrt{3} {e}^{- \sqrt{3} x}$
$\therefore y ' ' = 3 A {e}^{\sqrt{3} x} + 3 B {e}^{- \sqrt{3} x}$

Hence;

$y ' ' - 3 y = 3 A {e}^{\sqrt{3} x} + 3 B {e}^{- \sqrt{3} x} - 3 \left\{A \sqrt{3} {e}^{\sqrt{3} x} + B \sqrt{3} {e}^{- \sqrt{3} x}\right\} = 0$
-+-+-+-+-+-+-+-+-+-+-+-+-+-+