en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+2y=g(t),y(0)=8,y^'(0)=6

Solution

y^{^{''}} + 2 y^{^{'}} + 2 y = g (t), y (0) = 8, y^{^{'}} (0) = 6

Solution

y = e^{- t} ((8 + \frac{g}{2}) cos (t) + 14 sin (t)) + \frac{g t}{2} - \frac{g}{2}

Solution steps

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

Solve linear ODE:

y = e^{- t} ((8 + \frac{g}{2}) cos (t) + 14 sin (t)) + \frac{g t}{2} - \frac{g}{2}

y = e^{- t} ((8 + \frac{g}{2}) cos (t) + 14 sin (t)) + \frac{g t}{2} - \frac{g}{2}

Graph

Popular Examples

y^{''}-4y=1,y^'(0)=2

y^{^{''}} - 4 y = 1, y^{^{'}} (0) = 2

y^{''}-16y=19.2e^{4x}+60e^x

y^{^{''}} - 16 y = 19.2 e^{4 x} + 60 e^{x}

y^{''}-2y^'+5y=2(cos^2(x))e^x

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 (cos^{2} (x)) e^{x}

x^{''}-4x^'+4x=4e^{2t},x^0=-1,x^'0=-4

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 e^{2 t}, x^{0} = - 1, x^{^{'}} 0 = - 4

y^{''}-12y^'+36y=10e^{6x}

y^{^{''}} - 12 y^{^{'}} + 36 y = 10 e^{6 x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+2y^'+2y=g(t),y(0)=8,y^'(0)=6 ?
The solution for y^{''}+2y^'+2y=g(t),y(0)=8,y^'(0)=6 is y=e^{-t}((8+g/2)cos(t)+14sin(t))+(gt)/2-g/2