What is the solution for (D^2-D)y=e^xcos(x) ?

The solution for (D^2-D)y=e^xcos(x) is y=c_{1}e^x+c_{2}+1/2 e^xsin(x)-1/2 e^xcos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(D^2-D)y=e^xcos(x)

(D^{2} - D) y = e^{x} cos (x)

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

Solution steps

(D^{2} - D) y = e^{x} cos (x)

Solve linear ODE:

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

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 5 sin (x))

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = x e^{\frac{x}{2}}

4 x^{^{''}} - 2 x = 7

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

What is the solution for (D^2-D)y=e^xcos(x) ?
The solution for (D^2-D)y=e^xcos(x) is y=c_{1}e^x+c_{2}+1/2 e^xsin(x)-1/2 e^xcos(x)