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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} = x e^{\frac{3 x}{4}}

(d - 1)^{2} y = (e^{2} x + sin^{2} (x))

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

4 y^{^{''}} - 4 y^{^{'}} + y = 64 e^{\frac{t}{2}}

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

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