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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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