What is the solution for y^{''}=4y,y^'(0)=1,y(0)=0 ?

The solution for y^{''}=4y,y^'(0)=1,y(0)=0 is y=(e^{2t})/4-1/4 e^{-2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}=4y,y^'(0)=1,y(0)=0

y^{^{''}} = 4 y, y^{^{'}} (0) = 1, y (0) = 0

y = \frac{e ^{2 t}}{4} - \frac{1}{4} e^{- 2 t}

Solution steps

y^{''} (t) = 4 y

Solve linear ODE:

y = \frac{e ^{2 t}}{4} - \frac{1}{4} e^{- 2 t}

y = \frac{e ^{2 t}}{4} - \frac{1}{4} e^{- 2 t}

y^{^{''}} - 18 y^{^{'}} + 81 y = t^{- 2} e^{9 t}

x \to 1 lim (x^{2} + 3 - 2)

\frac{d}{d x} (\frac{5 ^{x}}{cos ( x )})

x \to 5 lim (3 x - 7)

y = x^{4} - 4 x^{2}, y = x^{2} - 4

What is the solution for y^{''}=4y,y^'(0)=1,y(0)=0 ?
The solution for y^{''}=4y,y^'(0)=1,y(0)=0 is y=(e^{2t})/4-1/4 e^{-2t}