What is the solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 is y=(5t^4)/3-(20t^2)/3-40/9 cos(sqrt(3)t)+40/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+3y=5t^4,y(0)=0,y^'(0)=0

y^{^{''}} + 3 y = 5 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{5 t ^{4}}{3} - \frac{20 t ^{2}}{3} - \frac{40}{9} cos (3 t) + \frac{40}{9}

Solution steps

y^{''} (t) + 3 y = 5 t^{4}

Solve linear ODE:

y = \frac{5 t ^{4}}{3} - \frac{20 t ^{2}}{3} - \frac{40}{9} cos (3 t) + \frac{40}{9}

y = \frac{5 t ^{4}}{3} - \frac{20 t ^{2}}{3} - \frac{40}{9} cos (3 t) + \frac{40}{9}

What is the solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+3y=5t^4,y(0)=0,y^'(0)=0 is y=(5t^4)/3-(20t^2)/3-40/9 cos(sqrt(3)t)+40/9