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

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

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

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

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

Solution steps

y^{''} (t) - y^{'} (t) + cy (t) = 0

Solve linear ODE:

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

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

4 y^{^{''}} - 28 y^{^{'}} + 49 y = 0

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

y^{^{''}} + 4 y = 0, y^{^{'}} (0) = 5

\frac{d ^{2} f}{d x ^{2}} = - a^{2} f (x)

\frac{d ^{2} y}{d x ^{2}} - 36 y = 0

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