What is the solution for ty^'+(t+1)y=2te^{-t} ?

The solution for ty^'+(t+1)y=2te^{-t} is y= t/(e^t)+(c_{1})/(e^tt)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ty^'+(t+1)y=2te^{-t}

t y^{^{'}} + (t + 1) y = 2 t e^{- t}

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

Solution steps

t y^{'} (t) + (t + 1) y = 2 t e^{- t}

Solve linear ODE:

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

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

\frac{2 s + 12}{s ^{2} + 6 s + 18}

\int \frac{5 x ^{3} - 4 x}{1 - x ^{4}} d x

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

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

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

What is the solution for ty^'+(t+1)y=2te^{-t} ?
The solution for ty^'+(t+1)y=2te^{-t} is y= t/(e^t)+(c_{1})/(e^tt)