# What is the general solution of the differential equation? :  2y'' + 3y' -y =0

Jun 10, 2017

$y = A {e}^{\left(- \frac{3}{4} - \frac{\sqrt{17}}{4}\right) x} + B {e}^{\left(- \frac{3}{4} + \frac{\sqrt{17}}{4}\right) x}$

#### Explanation:

We have:

$2 y ' ' + 3 y ' - y = 0$

This is a second order linear Homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

$2 {m}^{2} + 3 m - 1 = 0$

This has two distinct real solutions:

${m}_{1} = - \frac{3}{4} - \frac{\sqrt{17}}{4}$ and ${m}_{2} = - \frac{3}{4} + \frac{\sqrt{17}}{4}$

And so the solution to the DE is;

$\setminus \setminus \setminus \setminus \setminus y = A {e}^{{m}_{1} x} + B {e}^{{m}_{2} x}$ Where $A , B$ are arbitrary constants
$\therefore y = A {e}^{\left(- \frac{3}{4} - \frac{\sqrt{17}}{4}\right) x} + B {e}^{\left(- \frac{3}{4} + \frac{\sqrt{17}}{4}\right) x}$

Note

The given solution:

$y = {y}_{1} = {e}^{- 2 x}$

is not actually a solution of the given DE, as:

$y ' \setminus \setminus = - 2 {e}^{- 2 x}$
$y ' ' = 4 {e}^{- 2 x}$

And

$2 y ' ' + 3 y ' - y = 8 {e}^{- 2 x} - 6 {e}^{- 2 x} - {e}^{- 2 x} \ne 0$