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

The solution for y^{''}-2y^'-3y=x is y=c_{1}e^{3x}+c_{2}e^{-x}-x/3+2/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x}{3} + \frac{2}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x}{3} + \frac{2}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{x}{3} + \frac{2}{9}

\frac{d y}{d t} = t y

6 - 52 x - \frac{1}{2}

\frac{\partial}{\partial x} (5 x^{3} y^{2} - x ln (y))

\int \frac{2}{x 1 - 4 ln ^{2} ( x )} d x

ln (e^{r x} + s)

What is the solution for y^{''}-2y^'-3y=x ?
The solution for y^{''}-2y^'-3y=x is y=c_{1}e^{3x}+c_{2}e^{-x}-x/3+2/9