Question

S how that y = c 1

e

x + c 2

e

2 x

is the general solution of y

′′ − 3 y

′ + 2y = 0 on any

interval, and find the particular solution for w hich y 0 = − 1 a n d y

′

( 0) = 1.

Answer 1

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

Characteristic (auxiliary) equation

$$r^2-3r+2=0$$

$$(r-1)(r-2)=0$$

$$r_1=1, r_2=2$$

The general solution of the differential equation is

$$y=c_1e^x+c_2e^{2x}, x\in \R$$

Given $y(0)=-1, y'(0)=1$

$$y(0)=c_1e^0+c_2e^{2(0)}$$

$$c_1+c_2=-1$$

$$y'=c_1e^x+2c_2e^{2x}$$

$$y'(0)=c_1e^0+2c_2e^{2(0)}$$

$$c_1+2c_2=1$$

Then

$$c_1=-1-c_2$$

$$c_2=2$$

$$c_1=-3, c_2=2$$

The particular solution is

$$y=-3e^x+2e^{2x}$$