What is the solution for y^{'''}-3y^{''}+4y^'-2y=e^x ?

The solution for y^{'''}-3y^{''}+4y^'-2y=e^x is y=c_{1}e^x+e^x(c_{2}cos(x)+c_{3}sin(x))+e^xx

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}-3y^{''}+4y^'-2y=e^x

y^{^{'''}} - 3 y^{^{''}} + 4 y^{^{'}} - 2 y = e^{x}

y = c_{1} e^{x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + e^{x} x

Solution steps

y^{'''} (x) - 3 y^{''} (x) + 4 y^{'} (x) - 2 y = e^{x}

Solve linear ODE:

y = c_{1} e^{x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + e^{x} x

y = c_{1} e^{x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + e^{x} x

y^{^{''}} - 6 y^{^{'}} + 9 y = 2, y^{^{'}} (0) = 17

(d^{2} - 1) y = x sin^{3} (x) + cos (x)

(d^{4} + 8 d^{2} + 16) \cdot y = - 2

(d^{2} + 3 d + 2) y = e^{2} x + sin (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = x^{2} e^{3} x

What is the solution for y^{'''}-3y^{''}+4y^'-2y=e^x ?
The solution for y^{'''}-3y^{''}+4y^'-2y=e^x is y=c_{1}e^x+e^x(c_{2}cos(x)+c_{3}sin(x))+e^xx