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

The solution for y^{''}+y=e^t+t^3,y(0)=2,y^'(0)=0 is y= 3/2 cos(t)+11/2 sin(t)+1/2 e^t+t^3-6t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=e^t+t^3,y(0)=2,y^'(0)=0

y^{^{''}} + y = e^{t} + t^{3}, y (0) = 2, y^{^{'}} (0) = 0

y = \frac{3}{2} cos (t) + \frac{11}{2} sin (t) + \frac{1}{2} e^{t} + t^{3} - 6 t

Solution steps

y^{''} (t) + y = e^{t} + t^{3}

Solve linear ODE:

y = \frac{3}{2} cos (t) + \frac{11}{2} sin (t) + \frac{1}{2} e^{t} + t^{3} - 6 t

y = \frac{3}{2} cos (t) + \frac{11}{2} sin (t) + \frac{1}{2} e^{t} + t^{3} - 6 t

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- 2 t}

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 e^{5 t}

y^{^{''}} + 2 y^{^{'}} + y = 7 e^{- t}

y^{^{''''}} + 4 y = 2 e^{3 x}

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 4 y = 2 t

What is the solution for y^{''}+y=e^t+t^3,y(0)=2,y^'(0)=0 ?
The solution for y^{''}+y=e^t+t^3,y(0)=2,y^'(0)=0 is y= 3/2 cos(t)+11/2 sin(t)+1/2 e^t+t^3-6t