What is the solution for y^{''}-8y^'+16y=0,y(0)=3,y^'(0)=0 ?

The solution for y^{''}-8y^'+16y=0,y(0)=3,y^'(0)=0 is y=3e^{4t}-12e^{4t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'+16y=0,y(0)=3,y^'(0)=0

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y = 3 e^{4 t} - 12 e^{4 t} t

Solution steps

y^{''} (t) - 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = 3 e^{4 t} - 12 e^{4 t} t

y = 3 e^{4 t} - 12 e^{4 t} t

\frac{d}{d x} (2 x^{2} + 7)

f (x) = (4 x + 5) (5 x - 1), at x = 3

\int 25 e^{- 5 x} sin (5 x) d x

\int - \frac{x}{1 - x ^{2}} d x

\frac{d}{d x} (x, (x))

What is the solution for y^{''}-8y^'+16y=0,y(0)=3,y^'(0)=0 ?
The solution for y^{''}-8y^'+16y=0,y(0)=3,y^'(0)=0 is y=3e^{4t}-12e^{4t}t