What is the solution for y^{''}+9y=5sin(3x),y(0)=2,y^'(0)=0 ?

The solution for y^{''}+9y=5sin(3x),y(0)=2,y^'(0)=0 is y=2cos(3x)+5/18 sin(3x)-(5xcos(3x))/6

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y^{''}+9y=5sin(3x),y(0)=2,y^'(0)=0

y^{^{''}} + 9 y = 5 sin (3 x), y (0) = 2, y^{^{'}} (0) = 0

y = 2 cos (3 x) + \frac{5}{18} sin (3 x) - \frac{5 x cos ( 3 x )}{6}

Solution steps

y^{''} (x) + 9 y = 5 sin (3 x)

Solve linear ODE:

y = 2 cos (3 x) + \frac{5}{18} sin (3 x) - \frac{5 x cos ( 3 x )}{6}

y = 2 cos (3 x) + \frac{5}{18} sin (3 x) - \frac{5 x cos ( 3 x )}{6}

y^{^{''}} + 25 y = \frac{1}{100} cos (5 x)

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

\frac{d ^{2} x}{d t ^{2}} - 6 \frac{d x}{d t} - 7 x = 6

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{3} e^{t}

y^{^{''}} - y^{^{'}} - 2 y = e^{x} cos (x)

What is the solution for y^{''}+9y=5sin(3x),y(0)=2,y^'(0)=0 ?
The solution for y^{''}+9y=5sin(3x),y(0)=2,y^'(0)=0 is y=2cos(3x)+5/18 sin(3x)-(5xcos(3x))/6