What is the solution for y^{''}+y=2,y(0)=4,y^'(0)=2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=2,y(0)=4,y^'(0)=2

y^{^{''}} + y = 2, y (0) = 4, y^{^{'}} (0) = 2

y = 2 cos (t) + 2 sin (t) + 2

Solution steps

y^{''} (t) + y = 2

Solve linear ODE:

y = 2 cos (t) + 2 sin (t) + 2

y = 2 cos (t) + 2 sin (t) + 2

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = cos^{2} (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{x ^{2} + 1}

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