What is the solution for y^{''}+6y^'+25y=0,y(0)=3,y^'(0)=-6 ?

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

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 6

y = e^{- 3 t} (3 cos (4 t) + \frac{3}{4} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (3 cos (4 t) + \frac{3}{4} sin (4 t))

y = e^{- 3 t} (3 cos (4 t) + \frac{3}{4} sin (4 t))

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

4 \frac{d ^{2} y}{d x ^{2}} + 36 y = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2}{3} x}

y^{^{'''}} - 6 y^{^{''}} + y^{^{'}} - 6 y = 0

2 y - 3 y^{^{''}} = 0

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