What is the solution for 4y^{''}-y=sin(x) ?

The solution for 4y^{''}-y=sin(x) is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-1/5 sin(x)

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4y^{''}-y=sin(x)

4 y^{^{''}} - y = sin (x)

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{1}{5} sin (x)

Solution steps

4 y^{''} (x) - y = sin (x)

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{1}{5} sin (x)

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- \frac{x}{2}} - \frac{1}{5} sin (x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 x + 7

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

y^{^{''}} - 5 y^{^{'}} + 4 y = cos (x)

y^{^{''}} + 9 y = 2 e^{3 t} + 2 cos (2 t)

y^{^{''}} - y^{^{'}} = 8 e x

What is the solution for 4y^{''}-y=sin(x) ?
The solution for 4y^{''}-y=sin(x) is y=c_{1}e^{x/2}+c_{2}e^{-x/2}-1/5 sin(x)