What is the solution for y^{''}+4y=cos(x),y(pi)=0,y(0)=0 ?

The solution for y^{''}+4y=cos(x),y(pi)=0,y(0)=0 is y= 1/3 cos(2x)+c_{2}sin(2x)+1/3 cos(x)

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y^{''}+4y=cos(x),y(pi)=0,y(0)=0

y^{^{''}} + 4 y = cos (x), y (π) = 0, y (0) = 0

y = \frac{1}{3} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} cos (x)

Solution steps

y^{''} (x) + 4 y = cos (x)

Solve linear ODE:

y = \frac{1}{3} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} cos (x)

y = \frac{1}{3} cos (2 x) + c_{2} sin (2 x) + \frac{1}{3} cos (x)

x^{^{''}} - 10 x^{^{'}} + 25 x = 4 t e^{5 t}

x^{^{''}} + 9 x = 2 sin (t)

y^{^{''}} + 2 y^{^{'}} = 2

y^{^{''}} + 2 y^{^{'}} = 7

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = (1 + e^{- x})

What is the solution for y^{''}+4y=cos(x),y(pi)=0,y(0)=0 ?
The solution for y^{''}+4y=cos(x),y(pi)=0,y(0)=0 is y= 1/3 cos(2x)+c_{2}sin(2x)+1/3 cos(x)