What is the solution for y^{''}-2y^'+y=20sin(x) ?

The solution for y^{''}-2y^'+y=20sin(x) is y=c_{1}e^x+c_{2}xe^x+10cos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'+y=20sin(x)

y^{^{''}} - 2 y^{^{'}} + y = 20 sin (x)

y = c_{1} e^{x} + c_{2} x e^{x} + 10 cos (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 20 sin (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 10 cos (x)

y = c_{1} e^{x} + c_{2} x e^{x} + 10 cos (x)

y^{^{''}} + 4 y = cos (3 x)

2 y^{^{''}} - 16 y + 4 = 0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} + 1

y^{^{''}} + y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 9

What is the solution for y^{''}-2y^'+y=20sin(x) ?
The solution for y^{''}-2y^'+y=20sin(x) is y=c_{1}e^x+c_{2}xe^x+10cos(x)