What is the solution for 25y+12y^'+4y^{''}=-2 ?

The solution for 25y+12y^'+4y^{''}=-2 is y=e^{-(3t)/2}(c_{1}cos(2t)+c_{2}sin(2t))-2/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

25y+12y^'+4y^{''}=-2

25 y + 12 y^{^{'}} + 4 y^{^{''}} = - 2

y = e^{- \frac{3 t}{2}} (c_{1} cos (2 t) + c_{2} sin (2 t)) - \frac{2}{25}

Solution steps

25 y + 12 y^{'} (t) + 4 y^{''} (t) = - 2

Solve linear ODE:

y = e^{- \frac{3 t}{2}} (c_{1} cos (2 t) + c_{2} sin (2 t)) - \frac{2}{25}

y = e^{- \frac{3 t}{2}} (c_{1} cos (2 t) + c_{2} sin (2 t)) - \frac{2}{25}

y^{^{'''}} - y^{^{'}} = 4 x^{2} + 5 x - 2

y^{^{''}} + 3 y^{^{'}} + 2 y = 1 + 4 x e^{- 3 x}

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = e^{t}

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + x, y (0) = 0, y^{^{'}} (0) = 4

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

What is the solution for 25y+12y^'+4y^{''}=-2 ?
The solution for 25y+12y^'+4y^{''}=-2 is y=e^{-(3t)/2}(c_{1}cos(2t)+c_{2}sin(2t))-2/25