Answer to Question #300350 in Differential Equations for raji

Question

Show that y= c₁e^x + c₂e^2x 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:

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

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

$y''= c_1e^x + 4c_2e^{2x}$

Then

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

The initial conditions gives

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

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

Finally

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