What is the solution for y^{''}+2y^'-3y=-20sin(x) ?

The solution for y^{''}+2y^'-3y=-20sin(x) is y=c_{1}e^x+c_{2}e^{-3x}+4sin(x)+2cos(x)

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y^{''}+2y^'-3y=-20sin(x)

y^{^{''}} + 2 y^{^{'}} - 3 y = - 20 sin (x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = - 20 sin (x)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 3 y = 3 e^{2 t}

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 t} ln (t)

y^{^{''}} - 7 y^{^{'}} + 6 y = x, y^{1} (0) = e x

y^{^{''}} - 8 y^{^{'}} - 9 y = 10, y (0) = 0, y^{^{'}} (0) = 4

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

What is the solution for y^{''}+2y^'-3y=-20sin(x) ?
The solution for y^{''}+2y^'-3y=-20sin(x) is y=c_{1}e^x+c_{2}e^{-3x}+4sin(x)+2cos(x)