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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 2 x \cdot sin (x)

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

y^{^{''}} - 5 y = (x - 1) sin (x) + (x + 1) cos (x)

y^{^{''}} + 9 y^{^{'}} + 8 y = 3 e^{5 x}

y^{^{''}} - 6 y^{^{'}} + 25 y = 2 sin (t) - cos (t)

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