What is the solution for y^{''}-y=sin(t),y(0)=1,y(0)=-1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=sin(t),y(0)=1,y(0)=-1

y^{^{''}} - y = sin (t), y (0) = 1, y (0) = - 1

y = (1 - c_{2}) e^{t} + c_{2} e^{- t} - \frac{1}{2} sin (t)

Solution steps

y^{''} (t) - y = sin (t)

Solve linear ODE:

y = (1 - c_{2}) e^{t} + c_{2} e^{- t} - \frac{1}{2} sin (t)

y = (1 - c_{2}) e^{t} + c_{2} e^{- t} - \frac{1}{2} sin (t)

y^{^{''}} + y = 6 e^{- 2 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 2 y = x^{2} sin (2 x)

y^{^{''}} + 16 y = g (t), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{\frac{3}{2}} e^{x}

y^{^{''}} + 16 y = g (t), y (0) = 0, y^{^{'}} (0) = 5

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