Answer to Question #236035 in Differential Equations for Joy

Question

y=C1e^2x + C2e^-x

Answer 1

$$y=C_1e^{2x}+C_2e^{-x}$$

$$ye^x=C_1e^{3x}+C_2$$

Differentiate both sides with respect to $x$

$$y'e^x+ye^x=3C_1e^{3x}+0$$

$$y'+y=3C_1e^{2x}+0$$

$$y'+y=3C_1e^{2x}$$

Differentiate both sides with respect to $x$

$$y''+y'=6C_1e^{2x}$$

Then

$$y''+y'=2(y'+y)$$

$$y''-y'-2y=0$$

Differential equation

$$y''-y'-2y=0$$