Question #219774
Solve the Initial Value problem
y"− 4y′ + 3y = 0,y( 0 )= 7,y′ (0) = 11
1
Expert's answer
2021-07-23T07:49:36-0400
y4y+3y=0y''-4y'+3y=0

The characteristic equation is


r24r+3=0r^2-4r+3=0

(r1)(r3)=0(r-1)(r-3)=0

r1=1,r2=3r_1=1, r_2=3

The general solution of the homogeneous differential equation is


y(t)=c1et+c2e3ty(t)=c_1e^t+c_2e^{3t}

y(0)=7:7=c1e0+c2e3(0)y(0)=7: 7=c_1e^0+c_2e^{3(0)}

c1+c2=7c_1+c_2=7

y(t)=c1et+3c2e3ty'(t)=c_1e^t+3c_2e^{3t}

y(0)=11:11=c1e0+3c2e3(0)y'(0)=11: 11=c_1e^0+3c_2e^{3(0)}


c1+3c2=11c_1+3c_2=11

Then


c1=5,c2=2c_1=5, c_2=2

Solution of the Initial Value problem is


y(t)=5et+2e3ty(t)=5e^t+2e^{3t}


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