What is the solution for 4y^{''}+16y=1 ?

The solution for 4y^{''}+16y=1 is y=c_{1}cos(2t)+c_{2}sin(2t)+1/16

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4y^{''}+16y=1

4 y^{^{''}} + 16 y = 1

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{16}

Solution steps

4 y^{''} (t) + 16 y = 1

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{16}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{1}{16}

y^{^{''}} + 4 y = e^{4 x} sin (3 x)

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = x^{3} e^{2 x}

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

y^{^{'''}} + 16 y^{^{'}} = e^{- π x}

What is the solution for 4y^{''}+16y=1 ?
The solution for 4y^{''}+16y=1 is y=c_{1}cos(2t)+c_{2}sin(2t)+1/16