What is the solution for y^{''}-4y^'+4y=0,y(0)=-3,y^'(0)=1 ?

The solution for y^{''}-4y^'+4y=0,y(0)=-3,y^'(0)=1 is y=-3e^{2t}+7e^{2t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=0,y(0)=-3,y^'(0)=1

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 1

y = - 3 e^{2 t} + 7 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = - 3 e^{2 t} + 7 e^{2 t} t

y = - 3 e^{2 t} + 7 e^{2 t} t

x \to 3 lim (cos (π x) + 6)

\int \frac{4 3 x}{3} d x

\frac{\partial}{\partial x} (110 e^{- 0.1 t})

\frac{d}{d x} (6 x^{2} + 6 x)

\frac{d}{d x} (2 e^{x} sin (x))

What is the solution for y^{''}-4y^'+4y=0,y(0)=-3,y^'(0)=1 ?
The solution for y^{''}-4y^'+4y=0,y(0)=-3,y^'(0)=1 is y=-3e^{2t}+7e^{2t}t