What is the solution for y^'+4y=3sin(2t) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+4y=3sin(2t)

y^{^{'}} + 4 y = 3 sin (2 t)

y = \frac{3 ( 2 sin ( 2 t ) - cos ( 2 t ) )}{10} + \frac{c _{1}}{e ^{4 t}}

Solution steps

y^{'} (t) + 4 y = 3 sin (2 t)

Solve linear ODE:

y = \frac{3 ( 2 sin ( 2 t ) - cos ( 2 t ) )}{10} + \frac{c _{1}}{e ^{4 t}}

y = \frac{3 ( 2 sin ( 2 t ) - cos ( 2 t ) )}{10} + \frac{c _{1}}{e ^{4 t}}

\frac{\partial}{\partial x} (3 x^{4} y^{2} + 2 x y^{2} - 5)

(- 1, 4), (- 3, - 2)

\int arctan (4 t) d t

\frac{\partial}{\partial y} (- y)

\frac{ln ( x )}{x ^{6}}

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