What is the solution for y^{'''}+49y^'=0 ?

The solution for y^{'''}+49y^'=0 is y=c_{1}+c_{2}cos(7t)+c_{3}sin(7t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}+49y^'=0

y^{^{'''}} + 49 y^{^{'}} = 0

y = c_{1} + c_{2} cos (7 t) + c_{3} sin (7 t)

Solution steps

y^{'''} (t) + 49 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} cos (7 t) + c_{3} sin (7 t)

y = c_{1} + c_{2} cos (7 t) + c_{3} sin (7 t)

e^{x^{5}}

\int \frac{1}{x ^{2} - 4 x + 3} d x

\int \frac{2 ^{x}}{1 - 4 ^{x}} d x

\frac{d}{d x} (8 x^{2} + 7)

f (x) = (2 x^{3} + 2) (x^{4} - 3 x)

What is the solution for y^{'''}+49y^'=0 ?
The solution for y^{'''}+49y^'=0 is y=c_{1}+c_{2}cos(7t)+c_{3}sin(7t)