What is the solution for y^'+6y=e^{4t} ?

The solution for y^'+6y=e^{4t} is y=(e^{4t})/(10)+(c_{1})/(e^{6t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+6y=e^{4t}

y^{^{'}} + 6 y = e^{4 t}

y = \frac{e ^{4 t}}{10} + \frac{c _{1}}{e ^{6 t}}

Solution steps

y^{'} (t) + 6 y = e^{4 t}

Solve linear ODE:

y = \frac{e ^{4 t}}{10} + \frac{c _{1}}{e ^{6 t}}

y = \frac{e ^{4 t}}{10} + \frac{c _{1}}{e ^{6 t}}

\frac{d}{d x} (sec^{5} (x^{2} - 1))

3 x^{2} + 1

f (x) = x^{\frac{1}{9}}

f (x) = 2 x (2 - 4 x)^{4}, at x = 1

What is the solution for y^'+6y=e^{4t} ?
The solution for y^'+6y=e^{4t} is y=(e^{4t})/(10)+(c_{1})/(e^{6t)}