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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Solution steps

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

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

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