What is the solution for y^{''}-8y^'+16y=8x+4 ?

The solution for y^{''}-8y^'+16y=8x+4 is y=c_{1}e^{4x}+c_{2}xe^{4x}+x/2+1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'+16y=8x+4

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 x + 4

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 8 x + 4

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} - 7 y = 3 e^{2 x} - 2 e^{3 x}

y^{^{'''}} - 4 y^{^{''}} + 13 y^{^{'}} = 104 e^{4 x}

y^{^{''}} + y^{^{'}} = 30, y (0) = - 94 t + 60 t^{2}

0.5 x^{^{''}} + 5 x^{^{'}} + 12 x = e^{- 4 t} (4 t - 2)

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{3} + 2 x^{2} + 5 x + 2

What is the solution for y^{''}-8y^'+16y=8x+4 ?
The solution for y^{''}-8y^'+16y=8x+4 is y=c_{1}e^{4x}+c_{2}xe^{4x}+x/2+1/2