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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 4 x^{^{'}} + 5 x = cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{2 x}, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{'''}} + 4 y^{^{'}} = t e^{- 2 t} + 3 cos (2 t)

y^{^{''}} - y^{^{'}} = 3 e^{t}

y^{^{'''}} - 3 \cdot y^{^{''}} + 2 \cdot y^{^{'}} = - 3 \cdot x^{2} + 6

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