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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 3, y (0) = 1

x^{^{''}} - x^{^{'}} - 6 x = 0

y^{^{''}} + y^{^{'}} + 5 y = 0, y (0) = 0

x^{^{''}} - 6 x^{^{'}} + 8 x = 0

2 y^{^{''}} + 2 y^{^{'}} - 2 y = 0

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