What is the solution for y^{''}+4y=2sec(2t) ?

The solution for y^{''}+4y=2sec(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+1/2 ln(cos(2t))cos(2t)+tsin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=2sec(2t)

y^{^{''}} + 4 y = 2 sec (2 t)

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

Solution steps

y^{''} (t) + 4 y = 2 sec (2 t)

Solve linear ODE:

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

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

4 x^{2} + 5 x, at x = 3

\frac{\partial}{\partial x} (e^{x y} + x + y)

\int tan^{5} (x) d x

\frac{d x}{d y} = \frac{x ^{2} y ^{2}}{1 + x}

\int_{0}^{2} \frac{t}{( 3 - t ) ^{2}} d t

What is the solution for y^{''}+4y=2sec(2t) ?
The solution for y^{''}+4y=2sec(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+1/2 ln(cos(2t))cos(2t)+tsin(2t)