What is the solution for y^{''}+4y=4t^2-4t+10,y(0)=0,y^'(0)=3 ?

The solution for y^{''}+4y=4t^2-4t+10,y(0)=0,y^'(0)=3 is y=-2cos(2t)+2sin(2t)+t^2-t+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=4t^2-4t+10,y(0)=0,y^'(0)=3

y^{^{''}} + 4 y = 4 t^{2} - 4 t + 10, y (0) = 0, y^{^{'}} (0) = 3

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

Solution steps

y^{''} (t) + 4 y = 4 t^{2} - 4 t + 10

Solve linear ODE:

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

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

y^{^{'''}} - 16 y^{^{'}} = t e^{4 t}

y^{^{''}} + 4 y = 8 t

y^{^{''}} + 4 y^{^{'}} + 6 y = 1 + e^{- 1}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + y = 11 e^{x}

y^{^{''}} + 36 y = 13 sec^{2} (6 t)

What is the solution for y^{''}+4y=4t^2-4t+10,y(0)=0,y^'(0)=3 ?
The solution for y^{''}+4y=4t^2-4t+10,y(0)=0,y^'(0)=3 is y=-2cos(2t)+2sin(2t)+t^2-t+2