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

The solution for y^'+2y=e^{(4t)} is y=(e^{4t})/6+(c_{1})/(e^{2t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{e ^{4 t}}{6} + \frac{c _{1}}{e ^{2 t}}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{4 t}}{6} + \frac{c _{1}}{e ^{2 t}}

y = \frac{e ^{4 t}}{6} + \frac{c _{1}}{e ^{2 t}}

\int (x^{2} - x) d x

\frac{d}{d x} (- 1.8 x^{2} + 14.4 x)

\frac{\partial}{\partial x} (- 4 (x + y^{6}))

(2 y + 1)^{3} - 24 x = - 3, (- 1, - 2)

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

What is the solution for y^'+2y=e^{(4t)} ?
The solution for y^'+2y=e^{(4t)} is y=(e^{4t})/6+(c_{1})/(e^{2t)}