What is the solution for y^{''}+b^2y=cos(2t),y(0)=1 ?

The solution for y^{''}+b^2y=cos(2t),y(0)=1 is y=(1-1/(b^2-4))cos(bt)+c_{2}sin(bt)+(cos(2t))/(b^2-4)

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y^{''}+b^2y=cos(2t),y(0)=1

y^{^{''}} + b^{2} y = cos (2 t), y (0) = 1

y = (1 - \frac{1}{b ^{2} - 4}) cos (b t) + c_{2} sin (b t) + \frac{cos ( 2 t )}{b ^{2} - 4}

Solution steps

y^{''} (t) + b^{2} y = cos (2 t)

Solve linear ODE:

y = (1 - \frac{1}{b ^{2} - 4}) cos (b t) + c_{2} sin (b t) + \frac{cos ( 2 t )}{b ^{2} - 4}

y = (1 - \frac{1}{b ^{2} - 4}) cos (b t) + c_{2} sin (b t) + \frac{cos ( 2 t )}{b ^{2} - 4}

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 t

y^{^{''}} + y = - 16 \cdot cos (3 x)

y^{^{''}} - y = \frac{e ^{x}}{x ^{2}}

y^{^{''}} - 9 y = 1, y^{^{'}} (0) = 10

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+b^2y=cos(2t),y(0)=1 ?
The solution for y^{''}+b^2y=cos(2t),y(0)=1 is y=(1-1/(b^2-4))cos(bt)+c_{2}sin(bt)+(cos(2t))/(b^2-4)