What is the solution for y^'+5y=t+e^{-3t} ?

The solution for y^'+5y=t+e^{-3t} is y=(5t-1)/(25)+1/(2e^{3t)}+(c_{1})/(e^{5t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+5y=t+e^{-3t}

y^{^{'}} + 5 y = t + e^{- 3 t}

y = \frac{5 t - 1}{25} + \frac{1}{2 e ^{3 t}} + \frac{c _{1}}{e ^{5 t}}

Solution steps

y^{'} (t) + 5 y = t + e^{- 3 t}

Solve linear ODE:

y = \frac{5 t - 1}{25} + \frac{1}{2 e ^{3 t}} + \frac{c _{1}}{e ^{5 t}}

y = \frac{5 t - 1}{25} + \frac{1}{2 e ^{3 t}} + \frac{c _{1}}{e ^{5 t}}

\int \frac{x ^{3}}{8} d x

f (x) = e^{r x}

\frac{x ^{6}}{x ^{2}}

f (x) = 5 + 4 x^{2} - 2 x^{3}, (2, 5)

\int \frac{1}{30 - x ^{2}} d x

What is the solution for y^'+5y=t+e^{-3t} ?
The solution for y^'+5y=t+e^{-3t} is y=(5t-1)/(25)+1/(2e^{3t)}+(c_{1})/(e^{5t)}