What is the solution for 64y^{''}-48y^'+34y=0 ?

The solution for 64y^{''}-48y^'+34y=0 is y=e^{(3t)/8}(c_{1}cos((5t)/8)+c_{2}sin((5t)/8))

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64y^{''}-48y^'+34y=0

64 y^{^{''}} - 48 y^{^{'}} + 34 y = 0

y = e^{\frac{3 t}{8}} (c_{1} cos (\frac{5 t}{8}) + c_{2} sin (\frac{5 t}{8}))

Solution steps

64 y^{''} (t) - 48 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{8}} (c_{1} cos (\frac{5 t}{8}) + c_{2} sin (\frac{5 t}{8}))

y = e^{\frac{3 t}{8}} (c_{1} cos (\frac{5 t}{8}) + c_{2} sin (\frac{5 t}{8}))

\frac{d}{d x} (ln ((x^{2} + 1)^{20}))

\int \frac{x ^{2}}{( x + 1 ) ( x - 1 ) ^{2}} d x

\int_{1}^{2} x^{- 3} d x

t y^{^{'}} + (t + 1) y = t

\int 6 x (3 x^{2} - 2)^{4} d x

What is the solution for 64y^{''}-48y^'+34y=0 ?
The solution for 64y^{''}-48y^'+34y=0 is y=e^{(3t)/8}(c_{1}cos((5t)/8)+c_{2}sin((5t)/8))