What is the solution for 9y^{''}+4y=3x+12e^{-x} ?

The solution for 9y^{''}+4y=3x+12e^{-x} is y=c_{1}cos((2x)/3)+c_{2}sin((2x)/3)+(3x)/4+12/13 e^{-x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9y^{''}+4y=3x+12e^{-x}

9 y^{^{''}} + 4 y = 3 x + 12 e^{- x}

y = c_{1} cos (\frac{2 x}{3}) + c_{2} sin (\frac{2 x}{3}) + \frac{3 x}{4} + \frac{12}{13} e^{- x}

Solution steps

9 y^{''} (x) + 4 y = 3 x + 12 e^{- x}

Solve linear ODE:

y = c_{1} cos (\frac{2 x}{3}) + c_{2} sin (\frac{2 x}{3}) + \frac{3 x}{4} + \frac{12}{13} e^{- x}

y = c_{1} cos (\frac{2 x}{3}) + c_{2} sin (\frac{2 x}{3}) + \frac{3 x}{4} + \frac{12}{13} e^{- x}

y^{^{''''}} + 8 y^{^{''}} + 16 y = - sin (x)

x^{^{''}} + 6 x^{^{'}} + 23 x = 12 cos (4 t)

(d^{2} + 9) y = 30 sin^{3} (x)

y^{^{''''}} - y = 11 t + cos (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 e^{2 x}

What is the solution for 9y^{''}+4y=3x+12e^{-x} ?
The solution for 9y^{''}+4y=3x+12e^{-x} is y=c_{1}cos((2x)/3)+c_{2}sin((2x)/3)+(3x)/4+12/13 e^{-x}