What is the solution for y^{''}-y^'-by=0 ?

The solution for y^{''}-y^'-by=0 is y=c_{1}e^{(t(1+sqrt(1+4b)))/2}+c_{2}e^{(t(1-sqrt(1+4b)))/2}

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y^{''}-y^'-by=0

y^{^{''}} - y^{^{'}} - b y = 0

y = c_{1} e^{\frac{t ( 1 + 1 + 4 b )}{2}} + c_{2} e^{\frac{t ( 1 - 1 + 4 b )}{2}}

Solution steps

y^{''} (t) - y^{'} (t) - b y (t) = 0

Solve linear ODE:

y = c_{1} e^{\frac{t ( 1 + 1 + 4 b )}{2}} + c_{2} e^{\frac{t ( 1 - 1 + 4 b )}{2}}

y = c_{1} e^{\frac{t ( 1 + 1 + 4 b )}{2}} + c_{2} e^{\frac{t ( 1 - 1 + 4 b )}{2}}

y^{^{''}} + 9 y = 0, y^{^{'}} (π) = 0, y (0) = 0

y^{^{''}} + 4 y^{^{'}} + (3^{2} + 4^{2}) y = 0

y^{^{''}} + 2 y + y = 0, y^{1} (0) = x e^{- x}

3 y^{^{''}} - 12 y^{^{'}} - 36 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (\frac{1}{6}) = 2, y (0) = 3

What is the solution for y^{''}-y^'-by=0 ?
The solution for y^{''}-y^'-by=0 is y=c_{1}e^{(t(1+sqrt(1+4b)))/2}+c_{2}e^{(t(1-sqrt(1+4b)))/2}