What is the solution for t(dy)/(dt)+2y=t^{-3} ?

The solution for t(dy)/(dt)+2y=t^{-3} is y=-1/(t^3)+(c_{1})/(t^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t(dy)/(dt)+2y=t^{-3}

t \frac{d y}{d t} + 2 y = t^{- 3}

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

Solution steps

t \frac{d y}{d t} + 2 y = t^{- 3}

Solve linear ODE:

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

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

\int_{0}^{π} e^{x} sin (x) d x

x \to + 0 lim (\frac{sin ( x )}{4 x})

1500 x + 3000 (4 x + 3)^{- 1} - 1000

\frac{d}{d x} (f (x^{2})) = x^{4}

\frac{d ^{2} y}{d θ ^{2}} + y = 0

What is the solution for t(dy)/(dt)+2y=t^{-3} ?
The solution for t(dy)/(dt)+2y=t^{-3} is y=-1/(t^3)+(c_{1})/(t^2)