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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=2xsin(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

y^{^{''}} - y^{^{'}} = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5 x} ln (4 + x^{2})

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

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