What is the solution for 4y^{''}+24y^'+37y=0,y(pi)=1 ?

The solution for 4y^{''}+24y^'+37y=0,y(pi)=1 is y=e^{-3t}(c_{1}cos(t/2)+e^{3pi}sin(t/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}+24y^'+37y=0,y(pi)=1

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0, y (π) = 1

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

Solution steps

4 y^{''} (t) + 24 y^{'} (t) + 37 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 3 y^{^{''}} - 5 y^{^{'}} + 39 y = 0

y^{^{''}} + 29.4 y = 0

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

\frac{y ^{^{''}} + y + y}{x} = 0

y^{^{''}} + 2 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

What is the solution for 4y^{''}+24y^'+37y=0,y(pi)=1 ?
The solution for 4y^{''}+24y^'+37y=0,y(pi)=1 is y=e^{-3t}(c_{1}cos(t/2)+e^{3pi}sin(t/2))