What is the solution for y^{''}-4y^'-5y=0,y(1)=0 ?

The solution for y^{''}-4y^'-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 >

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

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

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

Solution steps

y^{''} (t) - 4 y^{'} (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^{^{''}} - 0.6 x^{^{'}} + 1.09 x = 0

((\frac{d ^{2} y}{d x ^{2}}) + 4 (\frac{d y}{d x}) + 20 y)^{2} = 0

y^{^{''}} + 16 y^{^{'}} + 63 y = 0

x^{^{''}} + 81 x = 0

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 26 y = 0

What is the solution for y^{''}-4y^'-5y=0,y(1)=0 ?
The solution for y^{''}-4y^'-5y=0,y(1)=0 is y=-c_{2}e^{5t-6}+c_{2}e^{-t}