What is the solution for y^{''}-y^'+30y=0 ?

The solution for y^{''}-y^'+30y=0 is y=e^{t/2}(c_{1}cos((sqrt(119)t)/2)+c_{2}sin((sqrt(119)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'+30y=0

y^{^{''}} - y^{^{'}} + 30 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{119 t}{2}) + c_{2} sin (\frac{119 t}{2}))

Solution steps

y^{''} (t) - y^{'} (t) + 30 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{119 t}{2}) + c_{2} sin (\frac{119 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{119 t}{2}) + c_{2} sin (\frac{119 t}{2}))

3 sin (2 t) - 3 cos (2 t)

\frac{d}{d x} (x^{y} - y^{x})

y^{^{'}} - y = x^{2}

x \to 7 lim (3 x - 5)

x \to 0 lim (\frac{2 x ^{2} + 5 x + 7}{x})

What is the solution for y^{''}-y^'+30y=0 ?
The solution for y^{''}-y^'+30y=0 is y=e^{t/2}(c_{1}cos((sqrt(119)t)/2)+c_{2}sin((sqrt(119)t)/2))