en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/4 y^{''}+2y^'+40y=-20/3 t-4/3

Solution

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 40 y = - \frac{20}{3} t - \frac{4}{3}

Solution

y = e^{- 4 t} (c_{1} cos (12 t) + c_{2} sin (12 t)) - \frac{t}{6} - \frac{1}{40}

Solution steps

\frac{1}{4} y^{''} (t) + 2 y^{'} (t) + 40 y = - \frac{20}{3} t - \frac{4}{3}

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (12 t) + c_{2} sin (12 t)) - \frac{t}{6} - \frac{1}{40}

y = e^{- 4 t} (c_{1} cos (12 t) + c_{2} sin (12 t)) - \frac{t}{6} - \frac{1}{40}

Graph

Popular Examples

y^{''}+2y^'+2y=10sin(2t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 10 sin (2 t)

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

y^{''}+3y=x^3-1

y^{^{''}} + 3 y = x^{3} - 1

y^{''}-8y^'+16y=8e^{2t},y(0)=3,y^'(0)=2

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{2 t}, y (0) = 3, y^{^{'}} (0) = 2

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

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

Frequently Asked Questions (FAQ)

What is the solution for 1/4 y^{''}+2y^'+40y=-20/3 t-4/3 ?
The solution for 1/4 y^{''}+2y^'+40y=-20/3 t-4/3 is y=e^{-4t}(c_{1}cos(12t)+c_{2}sin(12t))-t/6-1/40