What is the solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=-2 ?

The solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=-2 is y=e^{3t}(2cos(4t)-2sin(4t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=-2

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

y = e^{3 t} (2 cos (4 t) - 2 sin (4 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{3 t} (2 cos (4 t) - 2 sin (4 t))

y = e^{3 t} (2 cos (4 t) - 2 sin (4 t))

(4 D^{4} - 4 D^{3} - 23 D^{2} + 12 D + 36) y = 0

2500 y^{^{''}} - 9 y = 0, y (0) = 6, y^{^{'}} (0) = 7

\frac{d ^{3} x}{d t ^{3}} - x = 0

16 y^{^{''}} - 72 y^{^{'}} + 81 y = 0, y^{^{'}} (0) = 8

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

What is the solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=-2 ?
The solution for y^{''}-6y^'+25y=0,y(0)=2,y^'(0)=-2 is y=e^{3t}(2cos(4t)-2sin(4t))