Question #fcd5a

Nov 1, 2016

$y = - \frac{5}{3} {e}^{- 8 x} + \frac{14}{3} {e}^{2 x}$

Explanation:

This is a homogeneous linear differential equation with constant coefficients. The general solution to this kind of equation has the structure

$y \left(x\right) = {e}^{\lambda x}$. Substituting

$\left({\lambda}^{2} + 6 \lambda - 16\right) {e}^{\lambda x} = 0$. Here ${e}^{\lambda x} \ne 0$ so the equation is satisfied for $\lambda$'s satisfying

${\lambda}^{2} + 6 \lambda - 16 = 0$ or ${\lambda}_{1} = - 8 , {\lambda}_{2} = 2$
so the solution is

$y = {c}_{1} {e}^{- 8 x} + {c}_{2} {e}^{2 x}$ The constants ${c}_{1} , {c}_{2}$ are choosed according to the initial conditions. So

$y \left(0\right) = {c}_{1} + {c}_{2} = 3$ and
$y ' \left(0\right) = - 8 {c}_{1} + 2 {c}_{2} = - 4$ and finally, solving for ${c}_{1} , {c}_{2}$

${c}_{1} = - \frac{5}{3} , {c}_{2} = \frac{14}{3}$ Finally

$y = - \frac{5}{3} {e}^{- 8 x} + \frac{14}{3} {e}^{2 x}$