What is the solution for y^{''}-4y^'+3y=3x-1 ?

The solution for y^{''}-4y^'+3y=3x-1 is y=c_{1}e^{3x}+c_{2}e^x+x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+3y=3x-1

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 x - 1

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 3 x - 1

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 8 y = 3 e^{- 2 x} + 2 x

y^{^{''}} + 4 y = 8 cos (2 x) + 12 sin (2 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1

(D - 1)^{2} y = (e^{2 x} + sin (2 x))

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 3, y^{^{'}} (0) = 0

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