What is the solution for y^{''}-y^'-2y=8x+5 ?

The solution for y^{''}-y^'-2y=8x+5 is y=c_{1}e^{2x}+c_{2}e^{-x}-4x-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-2y=8x+5

y^{^{''}} - y^{^{'}} - 2 y = 8 x + 5

y = c_{1} e^{2 x} + c_{2} e^{- x} - 4 x - \frac{1}{2}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 8 x + 5

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - 4 x - \frac{1}{2}

y = c_{1} e^{2 x} + c_{2} e^{- x} - 4 x - \frac{1}{2}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{3 t}

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

y^{^{''}} - 8 y^{^{'}} - 4 y - 2 = 10 x^{2} - 2 x e^{2}

y^{^{''}} - y^{^{'}} - 6 y = 22 cos (3 x)

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

What is the solution for y^{''}-y^'-2y=8x+5 ?
The solution for y^{''}-y^'-2y=8x+5 is y=c_{1}e^{2x}+c_{2}e^{-x}-4x-1/2