What is the solution for y^{''}+2y^'+5y=45,y(0)=2,y^'(0)=3 ?

The solution for y^{''}+2y^'+5y=45,y(0)=2,y^'(0)=3 is y=e^{-t}(-7cos(2t)-2sin(2t))+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+5y=45,y(0)=2,y^'(0)=3

y^{^{''}} + 2 y^{^{'}} + 5 y = 45, y (0) = 2, y^{^{'}} (0) = 3

y = e^{- t} (- 7 cos (2 t) - 2 sin (2 t)) + 9

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 45

Solve linear ODE:

y = e^{- t} (- 7 cos (2 t) - 2 sin (2 t)) + 9

y = e^{- t} (- 7 cos (2 t) - 2 sin (2 t)) + 9

y^{^{''}} + y^{^{'}} - 20 y = e^{x} (82 + 84 x - 36 x^{2})

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 8 y = H (t)

(D - 1) (D - 2) (D - 3) y = 6 e^{4 x}

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

y^{^{''}} + g^{2} y = cos (2 t), y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'+5y=45,y(0)=2,y^'(0)=3 ?
The solution for y^{''}+2y^'+5y=45,y(0)=2,y^'(0)=3 is y=e^{-t}(-7cos(2t)-2sin(2t))+9