What is the solution for y^'+y=5cos(2t) ?

The solution for y^'+y=5cos(2t) is y=2sin(2t)+cos(2t)+(c_{1})/(e^t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+y=5cos(2t)

y^{^{'}} + y = 5 cos (2 t)

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

Solution steps

y^{'} (t) + y = 5 cos (2 t)

Solve linear ODE:

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

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

\frac{( 3 s + 7 )}{s ^{2} + 4 s + 3}

\frac{d}{d x} (x^{4} \cdot sin (x) \cdot sin (3 x))

\frac{\partial}{\partial t} (s^{2} + t^{2})

\int \frac{10}{( 20 x + 1 ) ^{23}} d x

\int 6 csc (x) cot (x) d x

What is the solution for y^'+y=5cos(2t) ?
The solution for y^'+y=5cos(2t) is y=2sin(2t)+cos(2t)+(c_{1})/(e^t)