What is the solution for (D^2+2D-3)y=cos(2x) ?

The solution for (D^2+2D-3)y=cos(2x) is y=c_{1}e^x+c_{2}e^{-3x}+4/65 sin(2x)-7/65 cos(2x)

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(D^2+2D-3)y=cos(2x)

(D^{2} + 2 D - 3) y = cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

Solution steps

(D^{2} + 2 D - 3) y = cos (2 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{4}{65} sin (2 x) - \frac{7}{65} cos (2 x)

y^{^{''}} - 8 y^{^{'}} = 6 e^{2 x} - 6 x^{2} - 9 x - 14

y^{^{''}} + 10 y^{^{'}} + 41 = 0

y^{^{''}} - 3 y^{^{'}} + 4 y = 16 x - 50 cos (2 x)

y^{^{''}} - y^{^{'}} - 3 y - 4 = 0

x^{^{''}} + 0.2 x^{^{'}} - x = 0.3 cos (1 t)

What is the solution for (D^2+2D-3)y=cos(2x) ?
The solution for (D^2+2D-3)y=cos(2x) is y=c_{1}e^x+c_{2}e^{-3x}+4/65 sin(2x)-7/65 cos(2x)