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}

\frac{\partial}{\partial x} (y^{2 x})

d y - (y - 1)^{2} d x = 0

x \to 1 lim (2 x^{3} - 5 x^{2} + x + 6)

\int cot^{3} (x) csc^{3} (x) d x

\frac{d}{d x} ((x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1))

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}