What is the solution for y^{''}+y^'-6y=sin(x) ?

The solution for y^{''}+y^'-6y=sin(x) is y=c_{1}e^{2x}+c_{2}e^{-3x}-7/50 sin(x)-1/50 cos(x)

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y^{''}+y^'-6y=sin(x)

y^{^{''}} + y^{^{'}} - 6 y = sin (x)

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{7}{50} sin (x) - \frac{1}{50} cos (x)

Solution steps

y^{''} (x) + y^{'} (x) - 6 y = sin (x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{7}{50} sin (x) - \frac{1}{50} cos (x)

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} - \frac{7}{50} sin (x) - \frac{1}{50} cos (x)

y^{^{''}} - 5 y + 6 y = e^{3 x} - x^{2}

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

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

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 11 t

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 12 t

What is the solution for y^{''}+y^'-6y=sin(x) ?
The solution for y^{''}+y^'-6y=sin(x) is y=c_{1}e^{2x}+c_{2}e^{-3x}-7/50 sin(x)-1/50 cos(x)