What is the solution for y^{''}+y=xsin(x),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+y=xsin(x),y(0)=1,y^'(0)=0 is y=-(x^2cos(x))/4+(xsin(x))/4+cos(x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=xsin(x),y(0)=1,y^'(0)=0

y^{^{''}} + y = x sin (x), y (0) = 1, y^{^{'}} (0) = 0

y = - \frac{x ^{2} cos ( x )}{4} + \frac{x sin ( x )}{4} + cos (x)

Solution steps

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

Solve linear ODE:

y = - \frac{x ^{2} cos ( x )}{4} + \frac{x sin ( x )}{4} + cos (x)

y = - \frac{x ^{2} cos ( x )}{4} + \frac{x sin ( x )}{4} + cos (x)

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = - 8 e x + 6 e - x

y^{^{''}} + 16 y = csc (4 x)

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

y^{^{''}} - y^{^{'}} = - 9

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

What is the solution for y^{''}+y=xsin(x),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+y=xsin(x),y(0)=1,y^'(0)=0 is y=-(x^2cos(x))/4+(xsin(x))/4+cos(x)