What is the solution for (d^2y)/(dt^2)-4(dy)/(dt)-5y=0,y(1)=0 ?

The solution for (d^2y)/(dt^2)-4(dy)/(dt)-5y=0,y(1)=0 is y=-c_{2}e^{5t-6}+c_{2}e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(d^2y)/(dt^2)-4(dy)/(dt)-5y=0,y(1)=0

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} - 5 y = 0, y (1) = 0

y = - c_{2} e^{5 t - 6} + c_{2} e^{- t}

Solution steps

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} - 5 y = 0

Solve linear ODE:

y = - c_{2} e^{5 t - 6} + c_{2} e^{- t}

y = - c_{2} e^{5 t - 6} + c_{2} e^{- t}

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

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

y^{^{''}} + 6 y^{^{'}} + 7 y = 0, y (0) = 2, y^{^{'}} (0) = 0

- y^{^{''}} - 9 \cdot π^{2} y = 0

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 324 x = 0

What is the solution for (d^2y)/(dt^2)-4(dy)/(dt)-5y=0,y(1)=0 ?
The solution for (d^2y)/(dt^2)-4(dy)/(dt)-5y=0,y(1)=0 is y=-c_{2}e^{5t-6}+c_{2}e^{-t}