What is the solution for y^{''}-8y^'+20y=0 ?

The solution for y^{''}-8y^'+20y=0 is y=e^{4t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}-8y^'+20y=0

y^{^{''}} - 8 y^{^{'}} + 20 y = 0

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

\int_{1}^{2} x 5^{(x^{2})} d x

x \to 0 lim (\frac{3 x ^{2} - 2 x}{x})

\int \frac{e ^{2 x} - 6 e ^{x}}{e ^{2 x} + 11} d x

t an g e n t f (x) = 20 x + 5

\int \frac{e ^{\frac{1}{x ^{4}}}}{x ^{5}} d x

What is the solution for y^{''}-8y^'+20y=0 ?
The solution for y^{''}-8y^'+20y=0 is y=e^{4t}(c_{1}cos(2t)+c_{2}sin(2t))