What is the solution for y^'+4/3 y=1-1/4 t ?

The solution for y^'+4/3 y=1-1/4 t is y=(3(-4t+19)}{64}+(c_{1})/(e^{\frac{4t)/3)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+4/3 y=1-1/4 t

y^{^{'}} + \frac{4}{3} y = 1 - \frac{1}{4} t

y = \frac{3 ( - 4 t + 19 )}{64} + \frac{c _{1}}{e ^{\frac{4 t}{3}}}

Solution steps

y^{'} (t) + \frac{4}{3} y = 1 - \frac{1}{4} t

Solve linear ODE:

y = \frac{3 ( - 4 t + 19 )}{64} + \frac{c _{1}}{e ^{\frac{4 t}{3}}}

y = \frac{3 ( - 4 t + 19 )}{64} + \frac{c _{1}}{e ^{\frac{4 t}{3}}}

\int_{- 1}^{0} 3 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

\frac{d}{d x} (\frac{1}{π} sin (π x))

x \to 5 lim (10 x + 35)

\frac{d}{d x} (ln (arcsin (x^{2})))

\int (x^{3} - 6 x) d x

What is the solution for y^'+4/3 y=1-1/4 t ?
The solution for y^'+4/3 y=1-1/4 t is y=(3(-4t+19)}{64}+(c_{1})/(e^{\frac{4t)/3)}