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

The solution for y^{'''}-3y^{''}+3y^'-y=x-2e^x is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x-x-3-(e^xx^3)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int 2 x + 4 x ln (8 x) d x

\frac{d}{d x} (1 + \frac{1}{3 x})

\int \frac{1}{sin ( 3 x )} d x

\int_{- 2}^{2} 3

(\frac{1}{x ^{2} - 1})^{^{'}}

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