Answer to Question #291593 in Differential Equations for kerthi

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

$y''−3y'+2y=0$

$D²y-3Dy+2y=0$

$(D²-3D+2)y=0$

This is a homogeneous equation. Thus, we have that

$m²-3m+2 = 0$

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

$=>m=2, 1$

The general solution is

$y=C_{1} e^x+C_{2}e^{2x}$

$y'=C_{1} e^x+2C_{2} e^{2x}$

But $y(0)=-1$ and $y'(0)=1$

$=>$

$C_1+C_2=-1$

$C_1+2C_2=1$

Solving both equations simultaneously, we have that

$∴C_1 = -3$ and $C_2 = 2$

Hence, the particular solution is

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