What is the solution for y^{''}+9y=3csc^2(3t) ?

The solution for y^{''}+9y=3csc^2(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(-ln(tan((3t)/2))cos(3t)-1)/(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+9y=3csc^2(3t)

y^{^{''}} + 9 y = 3 csc^{2} (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{- ln ( tan ( \frac{3 t}{2} ) ) cos ( 3 t ) - 1}{3}

Solution steps

y^{''} (t) + 9 y = 3 csc^{2} (3 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{- ln ( tan ( \frac{3 t}{2} ) ) cos ( 3 t ) - 1}{3}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{- ln ( tan ( \frac{3 t}{2} ) ) cos ( 3 t ) - 1}{3}

\int sinh (2 x) d x

2 x^{\frac{1}{5}} (x^{3} - 8)

x \to 0 lim (\frac{x}{cos ( x )})

\int t^{2} + t t - 1 + t sin (π) t d t

\int_{- 1}^{2} (x^{2} - 4 x + 3) d x

What is the solution for y^{''}+9y=3csc^2(3t) ?
The solution for y^{''}+9y=3csc^2(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(-ln(tan((3t)/2))cos(3t)-1)/(3)