What is the solution for y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=4 ?

The solution for y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=4 is y=2e^{5t}-6e^{5t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=4

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y = 2 e^{5 t} - 6 e^{5 t} t

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = 2 e^{5 t} - 6 e^{5 t} t

y = 2 e^{5 t} - 6 e^{5 t} t

y^{^{''}} - y^{^{'}} = 0

x^{^{''}} + 96 x = 0

y^{^{''}} - α y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = x e^{- x} ln (x)

25 y^{^{''}} + 80 y^{^{'}} + 64 y = 0, y^{^{'}} (0) = 5

What is the solution for y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=4 ?
The solution for y^{''}-10y^'+25y=0,y(0)=2,y^'(0)=4 is y=2e^{5t}-6e^{5t}t