Answer to Question #289483 in Differential Equations for ramya

Question

Show that 𝑦=𝑐1𝑒𝑥+𝑐2𝑒2𝑥 is the general solution of 𝑦′′−3𝑦′+2𝑦=0 on any interval, and find the particular solution for which 𝑦 0 =−1 and 𝑦′(0)=1. 

Answer 1

Given:

$𝑦=𝑐_1𝑒^𝑥+𝑐_2𝑒^{2𝑥}$

We have:

$y'=𝑐_1𝑒^𝑥+2𝑐_2𝑒^{2𝑥}$

$y''=𝑐_1𝑒^𝑥+4𝑐_2𝑒^{2𝑥}$

Hence

$$𝑦''−3𝑦'+2𝑦=(𝑐_1𝑒^𝑥+4𝑐_2𝑒^{2𝑥})\\ -3(𝑐_1𝑒^𝑥+2𝑐_2𝑒^{2𝑥})+2(𝑐_1𝑒^𝑥+𝑐_2𝑒^{2𝑥})=0$$

The initial conditions give

$$y(0)=c_1+c_2=-1$$

$$y'(0)=c_1+2c_2=1$$

Roots:

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

Finally

$$𝑦=-3𝑒^𝑥+2𝑒^{2𝑥}$$