What is the solution for y^{''}-4y^'+4y=0,y(0)=5,y^'=2 ?

The solution for y^{''}-4y^'+4y=0,y(0)=5,y^'=2 is y=5e^{2t}-8e^{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)=5,y^'=2

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = 2

y = 5 e^{2 t} - 8 e^{2 t} t

Solution steps

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

Solve linear ODE:

y = 5 e^{2 t} - 8 e^{2 t} t

y = 5 e^{2 t} - 8 e^{2 t} t

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

x^{^{''}} - 16 x = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 5

\frac{d ^{2} y}{d x ^{2}} - 13 \frac{d y}{d x} + 40 y = 0

(d^{2} - 2 d + 2) y = 0

What is the solution for y^{''}-4y^'+4y=0,y(0)=5,y^'=2 ?
The solution for y^{''}-4y^'+4y=0,y(0)=5,y^'=2 is y=5e^{2t}-8e^{2t}t