What is the solution for y^{'''}+3y^'-2x=0 ?

The solution for y^{'''}+3y^'-2x=0 is y=c_{1}+c_{2}cos(sqrt(3)x)+c_{3}sin(sqrt(3)x)+(x^2)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}+3y^'-2x=0

y^{^{'''}} + 3 y^{^{'}} - 2 x = 0

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

Solution steps

y^{'''} (x) + 3 y^{'} (x) - 2 x = 0

Solve linear ODE:

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

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

(D^{2} + 4 D + 5) y = e^{(- x)} + 15 x

y^{^{''}} + y = 2 cos (x), y^{^{'}} (π) = 0, y (0) = 0

(D^{2} + 1) y = cos^{2} (x)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 9 y = x e^{- 3 x}

y^{^{''}} + 11 y^{^{'}} + 18 y = 8 H (t)

What is the solution for y^{'''}+3y^'-2x=0 ?
The solution for y^{'''}+3y^'-2x=0 is y=c_{1}+c_{2}cos(sqrt(3)x)+c_{3}sin(sqrt(3)x)+(x^2)/3