MAT244--2019F > Quiz-3

Tut 0301 - Quiz 3

(1/1)

Dang Tongbo:
2y'' + 4y' -4y = 0 , y(0) = 0, y'(0) = 1.
Solution:
 2(r^2)+r-4= 0
so, r1 = ((-1)+(33)^(1/2))/4
     r2 = ((-1)+(33)^(1/2))/4
     y = c1*e^(((-1)+(33)^(1/2))/4)+c2*e^(((-1)-(33)^(1/2))/4). 
Because y(0)=0, so, c1+c2 = 0.
y' = -((-1)-(33)^(1/2))/4)*c1*e^(((-1)+(33)^(1/2))/4)-((1+(33)^1/2))*c2*e^(((-1)-(33)^(1/2))/4)
Because y'(0) = 1, so 1 = -(1-(33)^(1/2)/4)*c1 - (1+(33)^(1/2))/4)*c2
c1 = -c2,
so, -(1-(33)^(1/2)/4)*(-c2) - (1+(33)^(1/2))/4)*c2=1
        c2 = -(4/(2*(33)^1/2))
       c1 = 4/(2*(33)^1/2)
so, y = 4/(2*(33)^1/2)*e^((-1-(33)^1/2))/4)*t - (4/2*(33)^(1/2))/4)*t

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