What is the solution for y^{''}+by^'+6y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+by^'+6y=0,y(0)=1,y^'(0)=0 is

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y^{''}+by^'+6y=0,y(0)=1,y^'(0)=0

y^{^{''}} + b y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = e^{\frac{t ( - b + b ^{2} - 24 )}{2}} (1 + \frac{b - b ^{2} - 24}{2 b ^{2} - 24}) - \frac{e ^{\frac{t ( - b - b ^{2} - 24 )}{2}} ( b - b ^{2} - 24 )}{2 b ^{2} - 24}

Solution steps

y^{''} (t) + b y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{\frac{t ( - b + b ^{2} - 24 )}{2}} (1 + \frac{b - b ^{2} - 24}{2 b ^{2} - 24}) - \frac{e ^{\frac{t ( - b - b ^{2} - 24 )}{2}} ( b - b ^{2} - 24 )}{2 b ^{2} - 24}

y = e^{\frac{t ( - b + b ^{2} - 24 )}{2}} (1 + \frac{b - b ^{2} - 24}{2 b ^{2} - 24}) - \frac{e ^{\frac{t ( - b - b ^{2} - 24 )}{2}} ( b - b ^{2} - 24 )}{2 b ^{2} - 24}

x^{^{''}} + 64 x^{^{'}} = 0

3 \frac{d ^{2} y}{d x ^{2}} - 14 \frac{d y}{d x} - 5 y = 0

9 y^{^{''}} - 16 y^{^{'}} = 0

y^{^{'}} + y^{^{''}} = y^{^{'''}}

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = 2, y (1) = 0

What is the solution for y^{''}+by^'+6y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+by^'+6y=0,y(0)=1,y^'(0)=0 is