What is the solution for y^{''}-2y^'+y=x^3+8x ?

The solution for y^{''}-2y^'+y=x^3+8x is y=c_{1}e^x+c_{2}xe^x+x^3+6x^2+26x+40

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=x^3+8x

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 8 x

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 26 x + 40

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = x^{3} + 8 x

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 26 x + 40

y = c_{1} e^{x} + c_{2} x e^{x} + x^{3} + 6 x^{2} + 26 x + 40

y^{^{''}} - 9 y = (x^{2} - 8) sin (3 x)

y^{^{''}} - 3 y^{^{'}} = sin (4 t)

y^{^{''}} + 4 y = 2 x, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + a y^{^{'}} + b y = x, y (0) = 0, y^{^{'}} (0) = 0

(D^{2} - 4 D + 3) y = 20 cos (x)

What is the solution for y^{''}-2y^'+y=x^3+8x ?
The solution for y^{''}-2y^'+y=x^3+8x is y=c_{1}e^x+c_{2}xe^x+x^3+6x^2+26x+40