What is the solution for y^{''}-2y^'-3y=4ex-6 ?

The solution for y^{''}-2y^'-3y=4ex-6 is y=c_{1}e^{3x}+c_{2}e^{-x}-(4ex)/3+(8e+18)/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-3y=4ex-6

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 6

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 18}{9}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 4 e x - 6

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 18}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{4 e x}{3} + \frac{8 e + 18}{9}

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

y^{^{''}} + k^{2} y = e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} + 2 y^{^{'}} - 3 y = - 9 cos (2 x) - 33 sin (2 x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 12 x + 6

What is the solution for y^{''}-2y^'-3y=4ex-6 ?
The solution for y^{''}-2y^'-3y=4ex-6 is y=c_{1}e^{3x}+c_{2}e^{-x}-(4ex)/3+(8e+18)/9