What is the solution for y^{''}-(2a-1)y^'+a(a-1)y=0 ?

The solution for y^{''}-(2a-1)y^'+a(a-1)y=0 is y=c_{1}e^{at}+c_{2}e^{(a-1)t}

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y^{''}-(2a-1)y^'+a(a-1)y=0

y^{^{''}} - (2 a - 1) y^{^{'}} + a (a - 1) y = 0

y = c_{1} e^{a t} + c_{2} e^{(a - 1) t}

Solution steps

y^{''} (t) - (2 a - 1) y^{'} (t) + a (a - 1) y = 0

Solve linear ODE:

y = c_{1} e^{a t} + c_{2} e^{(a - 1) t}

y = c_{1} e^{a t} + c_{2} e^{(a - 1) t}

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

33 y^{^{''''}} + 323 y^{^{''}} + 400 y = 0

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} + 4 y = 0

y^{^{'''}} + 7 y^{^{''}} - 10 y = 0

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

What is the solution for y^{''}-(2a-1)y^'+a(a-1)y=0 ?
The solution for y^{''}-(2a-1)y^'+a(a-1)y=0 is y=c_{1}e^{at}+c_{2}e^{(a-1)t}