In exercises 1-10, find the general solution. Give the real form. y\" - 13y\' +

Question

In exercises 1-10, find the general solution. Give the real form. y" - 13y' + 42y = 0. y" + 7y' + 3y = 0. y" - 3 y' + 8y = 0. y" - 12y = 0. y" + 12y = 0. y" - 3y' + 9/4 y = 0. 2y" + 3y' = 0. y" - y' - 30y = 0. y" - 4y' + 4y = 0. 5y" - 2y' + y = 0. In exercises 11-16, solve the given initial-value problem. y" - 5y' + 6y = 0, y(0) = 1, y'(0) = 1 y" + 2y' + y = 0, y(2) = 1, y'(2) = 2 y" + 1/4 y = 0, y(pi) = 1, y'(pi) = -1 y" - 2y' + 2y = 0, y(0) = -1, y'(0) = -1 y" + 4y' + 4y = 0, y(-1) = 2, y'(-1) = 1 y" - 2y' + 5y = 0, y(pi/2) = 0, y'(pi/2) = 2

Explanation / Answer

Sry i thought only one ... here are the questions u asked for

1) y''-13y'+42y=0 => m^2 - 13m+42 =0 => m = 6 or 7

y = C1 e^(6t) + C2 e^(7t)

2) y''+7y'+3y = 0 => m^2 +7m+3 = 0 => m = (7+sqrt(37))/2 or (7-sqrt(37))/2

y= C1 e^(m1) + C2 e^(m2) where m1 and m2 are roots caluculated above

5) y''+12y=0 => m^2 +12 = 0 => m = 2sqrt(3) i or -2sqrt(3) i

y = C1 cos(2sqrt(3)t) + C2 sin(2sqrt(3)t)

7) 2y''+3y'=0 => 2m^2 + 3m = 0 => m = 0 or -3/2

y = C1 + C2 e^(-1.5t)

11) y''-5y'+6y=0 and y'(0)=y(0)=1

=>m^2 -5m+6 = 0 => m= 2 or 3

y= C1 e^(2t) + C2 e^(3t)

y(0)=1 => C1 + C2 = 1

y'(0)=1 => 2C1 + 3C2 = 1

we get C2=-1 and C1 =2

therefore y=e^(2t) - e^(3t)

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