What is the solution for y^{''}+4y=3cos(2x) ?

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

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y^{''}+4y=3cos(2x)

y^{^{''}} + 4 y = 3 cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x sin ( 2 x )}{4}

Solution steps

y^{''} (x) + 4 y = 3 cos (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x sin ( 2 x )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{3 x sin ( 2 x )}{4}

x^{^{''}} + 8 x^{^{'}} + 16 x = 8 sin (4 t)

y^{^{'''}} + y^{^{''}} = 5 x^{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = (x + 1) e^{x}

y^{^{''}} - 4 y^{^{'}} + 2 y = 5 e^{x}

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