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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(D^{4} - 6 D^{3} + 5 D^{2}) y = 0

y^{^{'''}} - 24 y^{^{''}} + 192 y^{^{'}} - 512 y = 0

y^{^{''}} + 4 \cdot π \cdot \frac{y ^{^{'}}}{20} + 100 \cdot π^{2} \cdot y = 0

(d^{6} - 4 d^{4} + 4 d^{2}) \cdot y = 0

4 y^{^{'''}} - 12 y^{^{'}} + 9 y^{^{'}} = 0

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