Calculus 2 help - Initial Value Problem?

y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4

r^2+4r+6=0

This equation is of form ax^2+bx+c

a = 1 b = 4 c = 6

x=[-b+/-sqrt(b^2-4ac)]/2a]

x=[-4 +/-sqrt(4^2-4(1)(6)]/(2)(1)

discriminant is b^2-4ac =-8

i^2 = -1, so √i^2 = i

No real roots: The complex roots are

x=[-4 +i √(8)] / (2)(1)

x=[-4 -i √(8)] / (2)(1)

x= -2+√2 i

x = -2 -√2 i

y = C1 e^(-2x) cos(√2 x) + C2 e^(-2x) sin(√2 x)

y(0) = 2

2 = C1 e^(0) cos (0) + C2 e^(0) sin(0)

2 = C1 (1) + C2(0)

C1 = 2

y' = -2 C1 e^(-2x) cos(√2 x) - √2 C1 e^(-2x) sin(√2 x) -2 C2 e^(-2x) sin(√2 x) + √2 C2 e^(-2x) cos(√2 x)

y'(0) = 4

4 = -2C1 e^(0) cos(0) - √2 C1 e^(0) sin(0) -2 C2 e^(0) sin(0) + √2 C2 e^(0) cos(0)

4 = -2C1 - 0 - 0 + √2 C2

4 = -2(2) + √2 C2

8 = √2 C2

C2 = 8 / √2

C2 = 4√2

y = C1 e^(-2x) cos(√2 x) + C2 e^(-2x) sin(√2 x)

y = 2 e^(-2x) cos(√2 x) + 4√2 e^(-2x) sin(√2 x)

y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4

r² + 4r + 6 = 0

r = -2 +/-√-2 , no real solutions

r1 = -2 + i√2

r2 = -2 - i√2

y = A* e^(-2+i√2)x + B* e^(-2-i√2)x , You know the formula: y = e^(ix) = cosx + isinx

y = A*e^(-2x)(cos√2x + i sin√2x) + B*e^(-2x)(cos√2x - i sin√2x)

y = e^(-2x)[(A + B)*cos√2x + i (A - B)sin√2x]

y = e^(-2x)[ C1 cos√2x + C2 sin√2x]

y' = (-2)e^(-2x)[C1 cos√2x + C2 sin√2x] + e^(-2x)*[C1√2*(-sin√2x) + C2√2*(cos√2x)]

2 = y(0) = 1*[C1*1 + C2*0] => C1 = 2

4 = y'(0) = (-2)*1[C1 + 0] + 1*[0 + C2√2*1]

4 = -2C1 + C2√2 = -4 + C2√2 => C2 = 8/√2 = 4√2

Finally:

y = e^(-2x)*(2cos√2x + 4√2sin√2x)

==========================