y′′−4y′+3y=0 The characteristic equation is
r2−4r+3=0
(r−1)(r−3)=0
r1=1,r2=3 The general solution of the homogeneous differential equation is
y(t)=c1et+c2e3t
y(0)=7:7=c1e0+c2e3(0)
c1+c2=7
y′(t)=c1et+3c2e3t
y′(0)=11:11=c1e0+3c2e3(0)
c1+3c2=11 Then
c1=5,c2=2
Solution of the Initial Value problem is
y(t)=5et+2e3t
Comments