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

The solution for y^{''}-y=sin(t),<y(0)=0,y^'(0)=1 is y=c_{1}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)=0,y^'(0)=1

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

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

Solution steps

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

Solve linear ODE:

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

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

5 y^{^{''}} + y^{^{'}} = - 2 x, y (0) = 0, y^{^{'}} (0) = - 20

4 y^{^{''}} - 4 y^{^{'}} + y = e^{\frac{x}{2}}

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

y^{^{''}} - 4 y^{^{'}} + 13 (t) = 0

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

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