What is the solution for y^'+9y=t+e-8t ?

The solution for y^'+9y=t+e-8t is y=-(7(9t-1))/(81)+e/9+(c_{1})/(e^{9t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+9y=t+e-8t

y^{^{'}} + 9 y = t + e - 8 t

y = - \frac{7 ( 9 t - 1 )}{81} + \frac{e}{9} + \frac{c _{1}}{e ^{9 t}}

Solution steps

y^{'} (t) + 9 y = t + e - 8 t

Solve linear ODE:

y = - \frac{7 ( 9 t - 1 )}{81} + \frac{e}{9} + \frac{c _{1}}{e ^{9 t}}

y = - \frac{7 ( 9 t - 1 )}{81} + \frac{e}{9} + \frac{c _{1}}{e ^{9 t}}

\frac{d}{d x} (x,)

n = 8 \sum \infty e^{7 - 6 n}

x y^{^{'}} = 2 x^{2} y + 3 x^{3} e^{x^{2}}, y (0) = 5

- x + 2, x^{2} - 1, [0, 1]

t sin (3 t)

What is the solution for y^'+9y=t+e-8t ?
The solution for y^'+9y=t+e-8t is y=-(7(9t-1))/(81)+e/9+(c_{1})/(e^{9t)}