What is the solution for y^{''}-20y^'+100y=0 ?

The solution for y^{''}-20y^'+100y=0 is y=c_{1}e^{10t}+c_{2}te^{10t}

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y^{''}-20y^'+100y=0

y^{^{''}} - 20 y^{^{'}} + 100 y = 0

y = c_{1} e^{10 t} + c_{2} t e^{10 t}

Solution steps

y^{''} (t) - 20 y^{'} (t) + 100 y = 0

Solve linear ODE:

y = c_{1} e^{10 t} + c_{2} t e^{10 t}

y = c_{1} e^{10 t} + c_{2} t e^{10 t}

What is the solution for y^{''}-20y^'+100y=0 ?
The solution for y^{''}-20y^'+100y=0 is y=c_{1}e^{10t}+c_{2}te^{10t}