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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} sin (x) + \frac{1}{2} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} sin (x) + \frac{1}{2} cos (x)

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{1}{2} sin (x) + \frac{1}{2} cos (x)

\frac{d ^{2} y}{d x ^{2}} + 36 y = 288 sin (6 x) + 96 cos (6 x)

y^{^{''}} + y = sec (x) + tan (x)

y^{^{''}} = 4 + (y^{^{'}}) 2

y^{^{''}} - y = e^{2 x} \cdot e^{e^{x}}

y^{^{''}} + 9 y = 4 (\frac{u ^{5}}{3} π - H (\frac{13}{3}) π)

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