What is the solution for (D^3+4D)y=sin(2x) ?

The solution for (D^3+4D)y=sin(2x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)-(xsin(2x))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(D^3+4D)y=sin(2x)

(D^{3} + 4 D) y = sin (2 x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) - \frac{x sin ( 2 x )}{8}

Solution steps

(D^{3} + 4 D) y = sin (2 x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) - \frac{x sin ( 2 x )}{8}

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) - \frac{x sin ( 2 x )}{8}

y^{^{''}} - 4 y = x^{- 1} e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 4 y = cos (4 x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} 4 + 8 y = 8 x^{2} + 8 x + 1

y^{^{'''}} + y^{^{''}} - y^{^{'}} - y = \frac{e ^{x} - e ^{- x}}{2}

y^{^{''}} + 9 y = cos (4 x)

What is the solution for (D^3+4D)y=sin(2x) ?
The solution for (D^3+4D)y=sin(2x) is y=c_{1}+c_{2}cos(2x)+c_{3}sin(2x)-(xsin(2x))/8