What is the solution for 8y^{''}-8y^'+52y=0 ?

The solution for 8y^{''}-8y^'+52y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8y^{''}-8y^'+52y=0

8 y^{^{''}} - 8 y^{^{'}} + 52 y = 0

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

Solution steps

8 y^{''} (t) - 8 y^{'} (t) + 52 y = 0

Solve linear ODE:

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

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

f (x) = \frac{x ^{2} + 1}{x}

n = 1 \sum \infty \frac{5}{n ( n + 2 )}

\frac{d}{d x} (\frac{x x + 1}{x + 1})

y^{^{'}} + 6 y = e^{t}

\int \frac{5}{3 - 2 cos ( x )} d x

What is the solution for 8y^{''}-8y^'+52y=0 ?
The solution for 8y^{''}-8y^'+52y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))