What is the solution for 6y^{''}-3y=x^2-x+7 ?

The solution for 6y^{''}-3y=x^2-x+7 is y=c_{1}e^{x/(sqrt(2))}+c_{2}e^{-x/(sqrt(2))}-(x^2)/3+x/3-11/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6y^{''}-3y=x^2-x+7

6 y^{^{''}} - 3 y = x^{2} - x + 7

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{x ^{2}}{3} + \frac{x}{3} - \frac{11}{3}

Solution steps

6 y^{''} (x) - 3 y = x^{2} - x + 7

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{x ^{2}}{3} + \frac{x}{3} - \frac{11}{3}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{x ^{2}}{3} + \frac{x}{3} - \frac{11}{3}

y^{^{''}} + 4 y = x^{2} + 5 cos (2 x)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 4 x + 6

y^{^{''}} + y^{^{'}} = 3 x^{2}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - y = sin^{2} (x)

(D^{2} + 4) y = 8 cos^{2} (x)

What is the solution for 6y^{''}-3y=x^2-x+7 ?
The solution for 6y^{''}-3y=x^2-x+7 is y=c_{1}e^{x/(sqrt(2))}+c_{2}e^{-x/(sqrt(2))}-(x^2)/3+x/3-11/3