What is the solution for y^{''}+4y=20e^t,y(0)=0,y^'(0)=2 ?

The solution for y^{''}+4y=20e^t,y(0)=0,y^'(0)=2 is y=-4cos(2t)-sin(2t)+4e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=20e^t,y(0)=0,y^'(0)=2

y^{^{''}} + 4 y = 20 e^{t}, y (0) = 0, y^{^{'}} (0) = 2

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

Solution steps

y^{''} (t) + 4 y = 20 e^{t}

Solve linear ODE:

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

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

y^{^{''}} - y = 4 x + 3

y^{^{''}} - 15 y^{^{'}} + 54 y = 39 + 54 x

x^{^{''}} + 3 x^{^{'}} + 2 x = 1, x (0) = 0, x^{^{'}} (0) = 0

y^{^{''}} - 9 y = 54 t

y^{^{''}} + 10 y^{^{'}} + 25 y = \frac{e ^{- 5 x}}{x ^{6}}

What is the solution for y^{''}+4y=20e^t,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+4y=20e^t,y(0)=0,y^'(0)=2 is y=-4cos(2t)-sin(2t)+4e^t