What is the solution for y^{''}+y=2cos(x),y^'(pi)=0,y(0)=0 ?

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

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y^{''}+y=2cos(x),y^'(pi)=0,y(0)=0

y^{^{''}} + y = 2 cos (x), y^{^{'}} (π) = 0, y (0) = 0

y = x sin (x) - π sin (x)

Solution steps

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

Solve linear ODE:

y = x sin (x) - π sin (x)

y = x sin (x) - π sin (x)

(D^{2} + 1) y = cos^{2} (x)

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 9 y = x e^{- 3 x}

y^{^{''}} + 11 y^{^{'}} + 18 y = 8 H (t)

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

y^{^{''}} - y = t, y (0) = 0, y^{^{'}} (0) = 1

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