What is the solution for y^{''}+9y=2cos(3x),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+9y=2cos(3x),y(0)=1,y^'(0)=0 is y=(xsin(3x))/3+cos(3x)

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y^{''}+9y=2cos(3x),y(0)=1,y^'(0)=0

y^{^{''}} + 9 y = 2 cos (3 x), y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) + 9 y = 2 cos (3 x)

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x^{2} \cdot e^{- 2 y})

\frac{\partial}{\partial x} (3 y^{2} x)

\frac{\partial}{\partial x} (6 x^{7 y})

\frac{d}{d x} (sin (x + sin (7 x)))

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

What is the solution for y^{''}+9y=2cos(3x),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+9y=2cos(3x),y(0)=1,y^'(0)=0 is y=(xsin(3x))/3+cos(3x)