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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (- \frac{1}{x - 1})

\frac{d y}{d x} = 3 y x

\frac{\partial}{\partial x} (\frac{6 y}{x ^{2} + y ^{2} + 5})

x \to 0 - lim (lo g_{3} (x))

\int (x^{3} + 2)^{2} 3 x^{2} d x

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