What is the solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 ?

The solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+(e^xx^3)/6+x-13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{'''}-3y^{''}+3y^'-y=e^x-x+16

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x + 16

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} + x - 13

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 3 y^{'} (x) - y = e^{x} - x + 16

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} + x - 13

y = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x} + \frac{e ^{x} x ^{3}}{6} + x - 13

(d^{3} - 3 d + 2) \cdot y = 540 x^{3} \cdot e^{- x}

(d^{3} + 8) \cdot y = x^{4} + 2 x + 1 + 3 sin (x)

(d^{2} + d - 2) y = 1 + x - x^{2}

(d^{3} + 2 d^{2} - 3 d) \cdot y = x \cdot e^{3 x}

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

What is the solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 ?
The solution for y^{'''}-3y^{''}+3y^'-y=e^x-x+16 is y=c_{1}e^x+c_{2}xe^x+c_{3}x^2e^x+(e^xx^3)/6+x-13