What is the solution for y^{''}-4y+4y=e^{-t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y+4y=e^{-t},y(0)=0,y^'(0)=0 is y=e^{-t}+t-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y+4y=e^{-t},y(0)=0,y^'(0)=0

y^{^{''}} - 4 y + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y = e^{- t} + t - 1

Solution steps

y^{''} (t) - 4 y + 4 y = e^{- t}

Isolate

y^{''} (t) : y^{''} (t) = e^{- t}

Solve

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

Solve

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

Apply initial conditions:

y = e^{- t} + t - 1

y = e^{- t} + t - 1

x \to + 0 lim (3^{x} + 7)

(\frac{t ^{3}}{t ^{8} + 9})^{2}

\int_{0.5}^{1} 1.5 x^{2} d x

\int (7 x + 4 y) d y

\frac{d}{d x} (\frac{( x ^{4} + 1 )}{x ^{2} - 1})

What is the solution for y^{''}-4y+4y=e^{-t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y+4y=e^{-t},y(0)=0,y^'(0)=0 is y=e^{-t}+t-1