What is the solution for (D^3+4D)y=e^x+sin(x) ?

The solution for (D^3+4D)y=e^x+sin(x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+1/5 e^x-1/3 cos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(D^3+4D)y=e^x+sin(x)

(D^{3} + 4 D) y = e^{x} + sin (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{5} e^{x} - \frac{1}{3} cos (x)

Solution steps

(D^{3} + 4 D) y = e^{x} + sin (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{5} e^{x} - \frac{1}{3} cos (x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{5} e^{x} - \frac{1}{3} cos (x)

y^{^{''}} + y = - cos (2 t), y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} - 6 y^{^{'}} + 5 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + 9 y = 4 + 5 x

y^{^{''}} - 3 y^{^{'}} = 15

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x^{- 4}

What is the solution for (D^3+4D)y=e^x+sin(x) ?
The solution for (D^3+4D)y=e^x+sin(x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)+1/5 e^x-1/3 cos(x)