What is the solution for y^{'''}+y^{''}+8y^'-10y=0 ?

The solution for y^{'''}+y^{''}+8y^'-10y=0 is y=c_{1}e^t+e^{-t}(c_{2}cos(3t)+c_{3}sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}+y^{''}+8y^'-10y=0

y^{^{'''}} + y^{^{''}} + 8 y^{^{'}} - 10 y = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) + 8 y^{'} (t) - 10 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (\frac{1}{( x - 2 ) ^{2}})

\int \frac{x ^{2}}{2 x ^{3} - 1} - 2 d x

\int_{0}^{1} \int_{0}^{1} e^{2 v - 1} d v d u

\int cot (\frac{x}{3}) d x

\int \frac{6 x ^{2} - 5}{x} d x

What is the solution for y^{'''}+y^{''}+8y^'-10y=0 ?
The solution for y^{'''}+y^{''}+8y^'-10y=0 is y=c_{1}e^t+e^{-t}(c_{2}cos(3t)+c_{3}sin(3t))