What is the solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 ?

The solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 is y=-3cos(t)+3sin(t)+3e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0

y^{^{''}} + y = 6 e^{- t}, y^{^{'}} (0) = 0, y (0) = 0

y = - 3 cos (t) + 3 sin (t) + 3 e^{- t}

Solution steps

y^{''} (t) + y = 6 e^{- t}

Solve linear ODE:

y = - 3 cos (t) + 3 sin (t) + 3 e^{- t}

y = - 3 cos (t) + 3 sin (t) + 3 e^{- t}

y = ∣ 6 x ∣, y = x^{2} - 7

n = 1 \sum \infty \frac{n}{n}

- y

\frac{d}{d x} (20 cos (2 x))

\frac{d}{d x} (2 ln (3 - x))

What is the solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 ?
The solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 is y=-3cos(t)+3sin(t)+3e^{-t}