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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(x^{2} y + 3 y) y^{^{'}} = 2 x y^{2} + 8 x

\frac{d y}{d x} = e^{- 3 y}

\frac{d x}{d t} = t^{3} (1 - x), x (0) = 3

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

\int \frac{x}{( x - 2 ) ^{3}} d x

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