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

The solution for y^{'''}-4y^{''}-5y^'=0 is y=c_{1}+c_{2}e^{-t}+c_{3}e^{5t}

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^{^{'''}} - 4 y^{^{''}} - 5 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- t} + c_{3} e^{5 t}

Solution steps

y^{'''} (t) - 4 y^{''} (t) - 5 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{5 t}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{5 t}

x y = - 100

\int 10 x \cdot x d x

f (x) = 3 x - 8

\int \frac{( x ^{2} + x + 1 )}{( x + 1 ) ^{2} ( x + 2 )} d x

y = a csc^{2} (k x)

What is the solution for y^{'''}-4y^{''}-5y^'=0 ?
The solution for y^{'''}-4y^{''}-5y^'=0 is y=c_{1}+c_{2}e^{-t}+c_{3}e^{5t}