What is the solution for 4y^{''}-4y^'+10y=0 ?

The solution for 4y^{''}-4y^'+10y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}-4y^'+10y=0

4 y^{^{''}} - 4 y^{^{'}} + 10 y = 0

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

Solution steps

4 y^{''} (t) - 4 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

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

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

x \to 6 lim (\frac{x ^{2}}{x - 6})

\frac{d y}{d x} + y^{5} x = - 6 y

y^{^{''}} + 4 y^{^{'}} + 29 y = 0

What is the solution for 4y^{''}-4y^'+10y=0 ?
The solution for 4y^{''}-4y^'+10y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))